This deck explains what makes a limit form indeterminate rather than merely undefined, lists the seven indeterminate forms, and applies L'Hopital's Rule with its hypotheses checked every single time. It targets the four classic errors: applying the rule to a determinate form, using the quotient rule instead of differentiating the top and the bottom separately, skipping the re-check before a second application, and forgetting to exponentiate at the end of a log-then-limit problem.
Subject: Calculus I · 135 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Calculus I - Deck 13
When substitution hands you nonsense, differentiate your way out - but only after you check that you are allowed to.
Objectives
By the end of this deck you can:
Warm-up
Discussion prompt
Before we open Indeterminate Forms and L'Hopital's Rule: without looking back, what was the main idea of Linear Approximation, Differentials, and Newton's Method, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers local linearity and the linearization of a function at a point, then differentials and propagated measurement error, using concavity to decide whether an estimate comes out too high or too low, and Newton's Method together with its failure modes. It targets the classic errors: centering at a point whose value you do not know, confusing the differential with the true change, getting the over-or-under direction backwards, and iterating Newton's Method with the previous function value instead of the previous x-value.
Section
Part 1
Concept
Before any technique, try the obvious thing: substitute the value the variable is approaching.
\[ \lim_{x \to 3} \frac{x^2 - 1}{x + 2} = \frac{3^2 - 1}{3 + 2} = \frac{8}{5} \]
If the function is continuous at that point, substitution is the answer. Most limits are settled in this one line. L'Hopital's Rule only enters when substitution refuses to settle it.
Counterexample
Discussion prompt
Before any technique, try the obvious thing: substitute the value the variable is approaching.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
If the function is continuous at that point, substitution is the answer. Most limits are settled in this one line. L'Hopital's Rule only enters when substitution refuses to settle it.
Picture it
Animation
Shows: Step zero is always direct substitution — a rendered Manim animation.
Rendered with Manim.
Takeaway: Most limits never need L'Hopital at all.
Picture it
Animation
Shows: Sometimes you apply it more than once — a rendered Manim animation.
Rendered with Manim.
Takeaway: Recheck the form after every application.
Concept
Substitution has exactly three kinds of outcome, and they call for three different responses.
| Substitution gives | What it means | Example |
|---|---|---|
| A real number | That is the limit. Done. | (x + 1)/(x + 2) at x = 0 gives 1/2 |
| Nonzero over zero | Determinate: the size blows up, so no finite limit | 1/x as x approaches 0 from the right |
| Zero over zero | Indeterminate: no information yet | (sin x)/x as x approaches 0 |
Only the third row needs a new tool. The middle row already has its answer - it just is not a number.
Comparison
Comparison matrix
From Three things substitution can hand you: refill the What it means column from what you know. The rest of the table is as it appeared.
| Substitution gives | What it means | Example |
|---|---|---|
| A real number | That is the limit. Done. | (x + 1)/(x + 2) at x = 0 gives 1/2 |
| Nonzero over zero | Determinate: the size blows up, so no finite limit | 1/x as x approaches 0 from the right |
| Zero over zero | Indeterminate: no information yet | (sin x)/x as x approaches 0 |
Intuition
Picture the numerator and the denominator both sprinting toward zero. The limit is asking: which one gets there faster, and by how much?
If the top collapses much faster, the ratio dies out to zero. If the bottom collapses faster, the ratio blows up. If they shrink at comparable speed, the ratio settles on a finite number - the ratio of their speeds.
The form 0/0 reports only that a race is happening. It never reports the winner. That is exactly why we call it indeterminate.
Analogy
Discussion prompt
Explain Zero over zero is a race, not a number by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Picture the numerator and the denominator both sprinting toward zero. The limit is asking: which one gets there faster, and by how much?
Concept
indeterminate form — A shape a limit takes on substitution - like zero over zero - that is consistent with many different answers. It tells you your method failed, not what the limit is.
Undefined means there is no value. Indeterminate means there is not enough information yet. Two very different messages.
\[ \frac{0}{0} \;\text{is indeterminate} \qquad \frac{5}{0} \;\text{is not} \]
Explain it
Discussion prompt
Explain Indeterminate is not the same as undefined to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Undefined means there is no value. Indeterminate means there is not enough information yet. Two very different messages.
Picture it
Animation
Shows: L'Hopital's rule set beside the quotient rule to distinguish them.
These two get confused constantly.
Takeaway: L'Hopital differentiates the top and the bottom separately. It is not the quotient rule, and using one in place of the other gives nonsense.
Concept
Every limit below substitutes to zero over zero. Their answers have nothing in common.
| Limit as x approaches 0 | Form | Actual value |
|---|---|---|
| (3x)/x | 0/0 | 3 |
| (x*x)/x | 0/0 | 0 |
| x/(x*x) | 0/0 | grows without bound |
| (sin x)/x | 0/0 | 1 |
Four limits, one form, four answers. If the form determined the answer, this table could not exist. That is the whole proof that zero over zero is indeterminate.
Trade off
Comparison matrix
From Same form, wildly different answers: every row here is a choice with a cost. Fill the Form column, then say which row you would actually pick and what you give up for it.
| Limit as x approaches 0 | Form | Actual value |
|---|---|---|
| (3x)/x | 0/0 | 3 |
| (x*x)/x | 0/0 | 0 |
| x/(x*x) | 0/0 | grows without bound |
| (sin x)/x | 0/0 | 1 |
Picture it
Animation
Shows: Converting a difference form — a rendered Manim animation.
Rendered with Manim.
Takeaway: Combine into a single fraction and the form declares itself.
Picture it
Animation
Shows: Same form, wildly different answers — a rendered Manim animation.
Rendered with Manim.
Takeaway: All three are 0/0, and no two agree. That is what indeterminate means.
Concept
Not every combination involving zero or infinity is a mystery. Some are fully decided, and using L'Hopital on them is an error.
| Form | Verdict | Value |
|---|---|---|
| infinity plus infinity | determinate | grows without bound |
| infinity times infinity | determinate | grows without bound |
| nonzero divided by zero | determinate | size grows without bound |
| zero divided by nonzero | determinate | zero |
| zero raised to infinity | determinate | zero |
| infinity raised to infinity | determinate | grows without bound |
In each of these, both pieces push the answer the same way. There is no tug of war, so there is nothing to determine.
Discrimination
Sort into buckets
Sort these by Value, from memory, without looking back at Forms that look scary but are determinate. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Picture it
Animation
Shows: Forms that look scary but are determinate — a rendered Manim animation.
Rendered with Manim.
Takeaway: Only some forms are genuinely undecided. Learn which.
Concept
There are exactly seven. Memorize them - recognizing the form is the first move in every problem in this deck.
\[ \frac{0}{0} \qquad \frac{\infty}{\infty} \qquad 0 \cdot \infty \qquad \infty - \infty \]
\[ 1^{\infty} \qquad 0^{0} \qquad \infty^{0} \]
L'Hopital's Rule applies directly to only the first two. The other five must be rewritten into one of those two first.
Picture it
Animation
Shows: The remaining indeterminate forms and the instruction to rewrite them as quotients.
Only two forms are ready to use as they stand.
Takeaway: Products, differences and power forms must first be rewritten as a quotient. Only then is the rule applicable.
Intuition
A power is a tug of war between the base and the exponent. Take a base creeping toward one while the exponent runs to infinity.
The base being slightly bigger than one pushes the product up. The base being only slightly bigger than one pushes it back toward one. Whichever effect wins depends on the exact rates - so the form cannot decide.
\[ \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^{x} = e \qquad \lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^{x} = 1 \]
Same form, two different answers. Notice both bases approach one and both exponents run to infinity - only the rate differs.
Picture it
Animation
Shows: Why the power forms are indeterminate too — a rendered Manim animation.
Rendered with Manim.
Takeaway: The compound-interest limit lives exactly here.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The tempting shortcuts: anything over itself is one, or zero over anything is zero.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The other shortcut would give 0.
Read the form as a signal, not a value: it means do more work.
Why: The other shortcut would give 0. The true value is 5, since the x cancels: 5x over x equals 5 for every x other than 0.
Trap
The tempting shortcuts: anything over itself is one, or zero over anything is zero.
\[ \lim_{x \to 0} \frac{5x}{x} \;\overset{?}{=}\; \frac{0}{0} \;\overset{?}{=}\; 1 \]
Both shortcuts fail on the same problem
Why: The other shortcut would give 0. The true value is 5, since the x cancels: 5x over x equals 5 for every x other than 0.
Read the form as a signal, not a value: it means do more work.
\[ \lim_{x \to 0} \frac{5x}{x} = \lim_{x \to 0} 5 = 5 \]
Cancel first, then substitute
Why: A limit ignores the single point x equals 0, so cancelling the common factor is legal. Check at x equal to 0.001: 0.005 divided by 0.001 is exactly 5.
Notation
Annotate
From Trap: treating zero over zero as an answer — read this one piece at a time. What is each part doing?
On: \( \lim_{x \to 0} \frac{5x}{x} \;\overset{?}{=}\; \frac{0}{0} \;\overset{?}{=}\; 1 \)
Concept
Every problem in this deck starts the same way: substitute, then write down the form you got. In words, on your paper.
Students who skip this step are the ones who apply L'Hopital's Rule to a limit that never needed it - and get a confidently wrong answer.
Picture it
Animation
Shows: Converting a product form — a rendered Manim animation.
Rendered with Manim.
Takeaway: You get to choose which factor goes downstairs.
Picture it
Animation
Shows: The two indeterminate forms the rule accepts, with the check highlighted.
Check the form before differentiating anything.
Takeaway: The rule applies to zero over zero and infinity over infinity, and nothing else. Applying it to a determinate form produces a confidently wrong answer.
Section
Part 2
Concept
Suppose two functions both approach zero at the same place. Then the limit of their ratio equals the limit of the ratio of their derivatives.
\[ \text{If } \lim_{x \to a} f(x) = 0 \text{ and } \lim_{x \to a} g(x) = 0, \text{ then} \]
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
Read the right-hand side carefully. It is the derivative of the top over the derivative of the bottom - two separate derivatives, not the derivative of the quotient.
Concept
The rule works identically when both pieces grow without bound instead of shrinking to zero.
\[ \text{If } \lim_{x \to a} f(x) = \pm\infty \text{ and } \lim_{x \to a} g(x) = \pm\infty, \text{ then } \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
The signs do not have to match. One piece running to positive infinity and the other to negative infinity is still the infinity-over-infinity form.
And the target can be any of these: a finite number, a one-sided approach, or the variable growing without bound in either direction. The rule covers all of them.
Concept
L'Hopital's Rule is a conditional statement. Four conditions have to hold before the equals sign is honest.
In practice, condition one is where students go wrong and condition four is where the rule quietly refuses to help. We will hit both.
Picture it
Figure (svg): Two curves both passing through the same point on the horizontal axis, each shown with its dashed tangent line through that point; the steeper curve has the steeper tangent.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Both functions hit zero at the same point. Zoom in on that point far enough and each smooth curve becomes indistinguishable from its own tangent line.
Intuition
Both functions hit zero at the same point. Zoom in on that point far enough and each smooth curve becomes indistinguishable from its own tangent line.
Figure (svg): Two curves both passing through the same point on the horizontal axis, each shown with its dashed tangent line through that point; the steeper curve has the steeper tangent.
\[ f(x) \approx f'(a)(x-a), \qquad g(x) \approx g'(a)(x-a) \]
\[ \frac{f(x)}{g(x)} \approx \frac{f'(a)(x-a)}{g'(a)(x-a)} = \frac{f'(a)}{g'(a)} \]
The shared factor cancels and what survives is the ratio of the two slopes. That is the whole idea: when both quantities start at zero, their ratio is decided by how fast each one leaves zero.
Ranking
Put in order
Put the moves of Worked example: a rational limit into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Top: 4 minus 4 is 0. Bottom: 4 plus 2 minus 6 is 0.
Worked example
Evaluate this limit.
\[ \lim_{x \to 2} \frac{x^2 - 4}{x^2 + x - 6} \]
Substitute first and name the form
Why: Top: 4 minus 4 is 0. Bottom: 4 plus 2 minus 6 is 0. The form is zero over zero, so the limit is indeterminate and L'Hopital's Rule is on the table.
Confirm the hypotheses
Why: Both pieces are polynomials, so both are differentiable everywhere. The denominator's derivative is 2x plus 1, which equals 5 at the target - nonzero nearby.
Differentiate the top and the bottom separately
Why: This is not the quotient rule. Replace the numerator with its own derivative and the denominator with its own derivative.
\[ \lim_{x \to 2} \frac{x^2 - 4}{x^2 + x - 6} = \lim_{x \to 2} \frac{2x}{2x + 1} \]
Substitute into the new quotient
Why: The new quotient is continuous at the target, so substitution finishes the job.
\[ \frac{2(2)}{2(2) + 1} = \frac{4}{5} \]
Verify by factoring instead
Why: Factoring gives the same answer without the rule: the shared factor cancels and substitution gives 4 over 5. A value near the target agrees too - at x equal to 2.001 the original ratio is 0.80004.
\[ \frac{(x-2)(x+2)}{(x-2)(x+3)} = \frac{x+2}{x+3} \;\longrightarrow\; \frac{4}{5} = 0.8 \]
Pattern
Every single problem, in this order. Skipping step 1 is the number one source of wrong answers.
Picture it
Animation
Shows: L'Hopital settles the growth race — a rendered Manim animation.
Rendered with Manim.
Takeaway: Each ratio is infinity over infinity, and each is decided by one application.
Elimination
Eliminate the wrong options
What is the value of the limit of (1 minus cosine x) divided by x squared as x approaches 0?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Substitution gives zero over zero. One application gives sine x over 2x, which is still zero over zero. A second application gives cosine x over 2, which approaches one half. Numerically, at x equal to 0.01 the original ratio is 0.4999958.
Check
Work it on paper first. Substitute, name the form, and repeat as needed.
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \]
Check your understanding
What is the value of the limit of (1 minus cosine x) divided by x squared as x approaches 0?
Answer: A
Why: Substitution gives zero over zero. One application gives sine x over 2x, which is still zero over zero. A second application gives cosine x over 2, which approaches one half. Numerically, at x equal to 0.01 the original ratio is 0.4999958.
Step zero
Discussion prompt
Worked example: a logarithm at one — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Substitute and name the form
Answer:
Worked example
This limit is the derivative of the natural logarithm at one in disguise.
\[ \lim_{x \to 1} \frac{\ln x}{x - 1} \]
Substitute and name the form
Why: The natural log of 1 is 0, and 1 minus 1 is 0. The form is zero over zero, so the rule applies.
Differentiate top and bottom separately
Why: The derivative of the natural log of x is 1 over x; the derivative of x minus 1 is 1.
\[ \lim_{x \to 1} \frac{\ln x}{x-1} = \lim_{x \to 1} \frac{1/x}{1} = \lim_{x \to 1} \frac{1}{x} = 1 \]
Verify from both sides numerically
Why: At x equal to 1.01 the ratio is 0.00995033 divided by 0.01, which is 0.995033. At x equal to 0.99 it is negative 0.0100503 divided by negative 0.01, which is 1.00503. The two bracket 1.
| x | value of the ratio |
|---|---|
| 0.99 | 1.00503 |
| 1.01 | 0.99503 |
| limit | 1 |
Concept
Nothing stops you from applying the rule again to the new quotient. But the new quotient is a brand new limit, and it earns the rule only by being indeterminate itself.
So after every application: substitute, name the form, decide again. Every time. No exceptions.
The moment the form stops being indeterminate, you are done - just evaluate. Applying the rule one more time past that point produces a wrong answer, and it is a very common one.
Picture it
Animation
Shows: Recheck the form after every application — a rendered Manim animation.
Rendered with Manim.
Takeaway: Applying it once too often reintroduces an error.
Estimation
Predict first
This one needs the rule twice - and a form check between them.
Commit before you compute: what does Worked example: two applications come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with a table of nearby values
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The ratio is closing in on 0.5 from above as x shrinks, which matches the answer one half.
Worked example
This one needs the rule twice - and a form check between them.
\[ \lim_{x \to 0} \frac{e^{x} - x - 1}{x^2} \]
Substitute and name the form
Why: Top: 1 minus 0 minus 1 is 0. Bottom: 0. The form is zero over zero.
Apply the rule once
Why: The derivative of the top is the exponential minus 1; the derivative of the bottom is 2x.
\[ \lim_{x \to 0} \frac{e^{x} - x - 1}{x^2} = \lim_{x \to 0} \frac{e^{x} - 1}{2x} \]
Re-check the form before going again
Why: Top: 1 minus 1 is 0. Bottom: 0. Still zero over zero, so a second application is legal - and only because we checked.
Apply the rule a second time
Why: The derivative of the exponential minus 1 is the exponential; the derivative of 2x is 2.
\[ \lim_{x \to 0} \frac{e^{x} - 1}{2x} = \lim_{x \to 0} \frac{e^{x}}{2} = \frac{1}{2} \]
Verify with a table of nearby values
Why: The ratio is closing in on 0.5 from above as x shrinks, which matches the answer one half.
| x | value of the original ratio |
|---|---|
| 0.1 | 0.5170918 |
| 0.01 | 0.5016708 |
| 0.001 | 0.5001667 |
Picture it
Animation
Shows: Each line of the worked example "two applications", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The ratio is closing in on 0.5 from above as x shrinks, which matches the answer one half.
Fill the middle
Fill in the blanks
From Trap: one application too many — finish the line. Write what belongs on the right of the equals sign before you look.
\longrightarrow \frac\frac{1}{2}___ \;\longrightarrow\; \frac______ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. After the first pass the form was 0 over 1, a perfectly ordinary number.
Trap
The student applies the rule twice out of habit, without re-checking the form after the first pass.
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2 + x} \]
\[ \longrightarrow \frac{\sin x}{2x + 1} \;\longrightarrow\; \frac{\cos x}{2} = \frac{1}{2} \]
The second application was illegal, and the answer is wrong
Why: After the first pass the form was 0 over 1, a perfectly ordinary number. The rule did not apply, and using it anyway produced one half instead of the true value 0.
Substitute after every application and stop the moment the form is decided.
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2 + x} \;\longrightarrow\; \lim_{x \to 0} \frac{\sin x}{2x + 1} \]
\[ \frac{\sin 0}{2(0) + 1} = \frac{0}{1} = 0 \]
Stop here: the form is no longer indeterminate
Why: Zero over one is a number, so substitution finishes it. Check at x equal to 0.01: the original ratio is 0.0000499996 over 0.0101, which is about 0.00495 - heading to 0, not to one half.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The phrase "differentiate the quotient" is misheard as "use the quotient rule."
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It is zero over zero again; pushing it through gives one half.
Differentiate the numerator and the denominator as two separate functions.
Why: It is zero over zero again; pushing it through gives one half. The true answer is 1, so the quotient rule cost a factor of two.
Trap
The phrase "differentiate the quotient" is misheard as "use the quotient rule."
\[ \lim_{x \to 0} \frac{e^{x} - 1}{x} \;\overset{?}{=}\; \lim_{x \to 0} \frac{x e^{x} - (e^{x} - 1)}{x^2} \]
That new limit is a different problem entirely
Why: It is zero over zero again; pushing it through gives one half. The true answer is 1, so the quotient rule cost a factor of two.
\[ \lim_{x \to 0} \frac{x e^{x} - e^{x} + 1}{x^2} = \lim_{x \to 0} \frac{x e^{x}}{2x} = \frac{1}{2} \quad (\text{wrong}) \]
Differentiate the numerator and the denominator as two separate functions.
\[ \lim_{x \to 0} \frac{e^{x} - 1}{x} = \lim_{x \to 0} \frac{e^{x}}{1} = 1 \]
Verify numerically
Why: At x equal to 0.01 the original ratio is 0.010050167 divided by 0.01, which is 1.0050167 - right next to 1, nowhere near one half. This limit is also the definition of the derivative of the exponential at zero, which is 1.
Trap
Any limit that looks hard gets the rule applied on reflex.
\[ \lim_{x \to 0^{+}} \frac{x + 1}{x} \;\overset{?}{=}\; \lim_{x \to 0^{+}} \frac{1}{1} = 1 \]
The answer 1 is badly wrong
Why: Substitution gives 1 over 0, which is determinate: the numerator is heading to 1 while the denominator vanishes, so the ratio explodes. At x equal to 0.001 the true ratio is 1001.
Substitute first. A nonzero number over zero is already an answer.
\[ \lim_{x \to 0^{+}} \frac{x + 1}{x} = \infty \]
Check the size at a nearby point
Why: At x equal to 0.001 the ratio is 1.001 divided by 0.001, which is 1001; at x equal to 0.0001 it is 10001. The values grow without bound, confirming a vertical asymptote rather than the finite value 1.
Translation
\( \lim_{x \to 0^{+}} \frac{x + 1}{x} = \infty \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Prediction
Predict first
For which limit is L'Hopital's Rule NOT valid?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: The limit of cosine x over x squared as x approaches 0
Why: Substituting into C gives cosine of 0 over 0, which is 1 over 0 - determinate, and the true limit grows without bound. Applying the rule anyway gives negative sine x over 2x, which approaches negative one half: a finite, wrong, and even wrongly signed answer.
Check
Three of these four limits are fair game for L'Hopital's Rule. One is not. Substitute into each before you answer.
\[ \text{(A) } \lim_{x \to 0} \frac{\sin 3x}{5x} \quad \text{(B) } \lim_{x \to \infty} \frac{\ln x}{x} \quad \text{(C) } \lim_{x \to 0} \frac{\cos x}{x^2} \quad \text{(D) } \lim_{x \to 1} \frac{\ln x}{x-1} \]
Check your understanding
For which limit is L'Hopital's Rule NOT valid?
Answer: C
Why: Substituting into C gives cosine of 0 over 0, which is 1 over 0 - determinate, and the true limit grows without bound. Applying the rule anyway gives negative sine x over 2x, which approaches negative one half: a finite, wrong, and even wrongly signed answer.
Section
Part 3
Concept
The infinity-over-infinity form is the same race run in the opposite direction. Both pieces are growing without bound, and the question is which one grows faster.
If the bottom outruns the top, the ratio dies to zero. If the top outruns the bottom, the ratio grows without bound. If they keep pace, the ratio settles on the ratio of their rates.
This is why L'Hopital's Rule is the standard tool for comparing growth rates: differentiating repeatedly strips a polynomial down to a constant while an exponential is left completely unharmed.
Step zero
Discussion prompt
Worked example: a power against an exponential — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the form
Answer:
Worked example
Which grows faster, a squared term or the natural exponential?
\[ \lim_{x \to \infty} \frac{x^2}{e^{x}} \]
Name the form
Why: Both the numerator and the denominator grow without bound, so this is the infinity over infinity case - indeterminate, and the rule applies directly.
Apply the rule once
Why: The derivative of the squared term is 2x; the exponential is its own derivative, so the denominator is unchanged.
\[ \lim_{x \to \infty} \frac{x^2}{e^{x}} = \lim_{x \to \infty} \frac{2x}{e^{x}} \]
Re-check, then apply again
Why: The numerator 2x still grows without bound and so does the exponential, so the form is still infinity over infinity. One more pass flattens the numerator to a constant.
\[ \lim_{x \to \infty} \frac{2x}{e^{x}} = \lim_{x \to \infty} \frac{2}{e^{x}} = 0 \]
Stop: the form is now determinate
Why: The last quotient is a fixed 2 over something growing without bound, which is 2 divided by an enormous number - that is 0, not an indeterminate form.
Verify with actual numbers
Why: The ratio is already collapsing hard by the time x reaches 20, which matches a limit of 0.
| x | value of the ratio |
|---|---|
| 5 | 0.674 |
| 10 | 0.00454 |
| 20 | 0.000000824 |
Picture it
Animation
Shows: Each line of the worked example "a power against an exponential", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The ratio is already collapsing hard by the time x reaches 20, which matches a limit of 0.
Intuition
Think of each application of the rule as one round of a boxing match where both fighters take the same punch: a derivative.
A polynomial loses one degree per round. After enough rounds it is a constant, and one round later it is nothing. The natural exponential takes the punch and stands up completely unchanged, forever.
That is the entire reason exponentials beat every power, no matter how big the power is. A term raised to the one-hundredth power just means one hundred rounds - and the exponential is still standing.
Sorting
Sort into buckets
These are the pieces of Indeterminate Forms and L'Hopital's Rule, out of order. Put each one back under the part of the lesson it belongs to.
Concept
Every one of these comparisons is proved the same way: form the quotient and apply the rule until the form resolves.
| Rank | Family | Example | Beaten by |
|---|---|---|---|
| Slowest | logarithms | natural log of x | every positive power |
| Middle | roots and powers | square root of x, x, x cubed | every exponential |
| Fast | exponentials with base above 1 | 2 to the x, e to the x | the next row |
| Fastest | variable base and exponent | x to the x | nothing here |
\[ \lim_{x \to \infty} \frac{\ln x}{x^{p}} = 0 \quad (p > 0), \qquad \lim_{x \to \infty} \frac{x^{p}}{e^{x}} = 0 \quad (\text{any } p) \]
Memorize the ordering, but be ready to prove any one line of it on demand - that is a standard exam question.
Comparison
Comparison matrix
From The growth-rate hierarchy: refill the Family column from what you know. The rest of the table is as it appeared.
| Rank | Family | Example | Beaten by |
|---|---|---|---|
| Slowest | logarithms | natural log of x | every positive power |
| Middle | roots and powers | square root of x, x, x cubed | every exponential |
| Fast | exponentials with base above 1 | 2 to the x, e to the x | the next row |
| Fastest | variable base and exponent | x to the x | nothing here |
Estimation
Predict first
The square root grows slowly. Does the logarithm grow more slowly still?
Commit before you compute: what does Worked example: a logarithm against a root come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with large inputs
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The values keep shrinking as x grows by factors of one hundred, consistent with a limit of 0 - the logarithm loses to even a square root.
Worked example
The square root grows slowly. Does the logarithm grow more slowly still?
\[ \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} \]
Name the form
Why: Both the logarithm and the square root grow without bound, so this is infinity over infinity.
Rewrite the root as a power before differentiating
Why: A fractional exponent makes the power rule mechanical and prevents sign and placement errors.
\[ \frac{d}{dx}\left[x^{1/2}\right] = \tfrac{1}{2}x^{-1/2} \]
Apply the rule and simplify the complex fraction
Why: Dividing by one half of a negative power is the same as multiplying by twice the positive power, which collapses the whole thing.
\[ \lim_{x \to \infty} \frac{1/x}{\tfrac{1}{2}x^{-1/2}} = \lim_{x \to \infty} \frac{2x^{1/2}}{x} = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0 \]
Verify with large inputs
Why: The values keep shrinking as x grows by factors of one hundred, consistent with a limit of 0 - the logarithm loses to even a square root.
| x | value of the ratio |
|---|---|
| 100 | 0.4605 |
| 10000 | 0.0921 |
| 1000000 | 0.0138 |
Picture it
Animation
Shows: Each line of the worked example "a logarithm against a root", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The values keep shrinking as x grows by factors of one hundred, consistent with a limit of 0 - the logarithm loses to even a square root.
Ranking
Put in order
These are the steps of How to compare two growth rates, scrambled. Put them back in order before the next slide shows you.
Why: This is the order the recipe itself gives. Recalling the sequence without the slide in front of you is the difference between recognising the method and being able to run it — most of what goes wrong in practice is a step done out of turn.
Pattern
To decide which of two functions grows faster, do not argue - build the quotient and take a limit.
Check
How many applications does this need, and where does it land?
\[ \lim_{x \to \infty} \frac{x^{3}}{e^{x}} \]
Check your understanding
What is the limit of x cubed over the natural exponential as x grows without bound?
Answer: A
Why: Three applications turn the numerator into 3x squared, then 6x, then the constant 6, while the exponential in the denominator is unchanged each time. A fixed 6 over something growing without bound is 0.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The form is infinity over infinity, so the rule is legal - but it never finishes.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Differentiating once more flips the fraction over again and returns the original expression.
When the rule cycles, abandon it and divide by the highest power instead.
Why: Differentiating once more flips the fraction over again and returns the original expression. You can do this forever and learn nothing.
Trap
The form is infinity over infinity, so the rule is legal - but it never finishes.
\[ \lim_{x \to \infty} \frac{\sqrt{x^2 + 1}}{x} \;\longrightarrow\; \lim_{x \to \infty} \frac{x/\sqrt{x^2+1}}{1} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}} \]
Apply it again and you are back where you started
Why: Differentiating once more flips the fraction over again and returns the original expression. You can do this forever and learn nothing.
When the rule cycles, abandon it and divide by the highest power instead.
\[ \frac{\sqrt{x^2+1}}{x} = \frac{\sqrt{x^2+1}}{\sqrt{x^2}} = \sqrt{1 + \frac{1}{x^2}} \;\longrightarrow\; 1 \]
Verify at a large value
Why: At x equal to 1000 the expression equals the square root of 1.000001, which is about 1.0000005. The limit is 1, and no amount of differentiating would have told you that.
Break the constraint
Discussion prompt
The rule this trap just fixed:
At x equal to 1000 the expression equals the square root of 1.000001, which is about 1.0000005. The limit is 1, and no amount of differentiating would have told you that.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
Differentiating once more flips the fraction over again and returns the original expression. You can do this forever and learn nothing.
Concept
The rule being valid is not a reason to use it. For a ratio of polynomials, dividing by the highest power in the denominator is one line; L'Hopital's Rule can take several.
\[ \lim_{x \to \infty} \frac{3x^2 - 5}{2x^2 + 7} = \lim_{x \to \infty} \frac{3 - 5/x^2}{2 + 7/x^2} = \frac{3}{2} \]
The rule would need two applications to reach the same place. Reach for it when algebra has run out, not before.
Section
Part 4
Concept
L'Hopital's Rule is a statement about a fraction. A product, a difference, or a power is not a fraction, so the rule cannot touch it directly.
The whole job with the other five forms is translation: rewrite the expression as a quotient that is zero over zero or infinity over infinity, then run the procedure you already know.
| Form | Move that turns it into a quotient |
|---|---|
| zero times infinity | send one factor to the denominator as its reciprocal |
| infinity minus infinity | combine over a common denominator, or multiply by a conjugate |
| one to the infinity | take the natural log, find that limit, then exponentiate |
| zero to the zero | take the natural log, find that limit, then exponentiate |
| infinity to the zero | take the natural log, find that limit, then exponentiate |
Concept
A product where one factor dies to zero and the other explodes is a tug of war with no fraction bar in sight. Give it one.
\[ f(x)\,g(x) = \frac{f(x)}{1/g(x)} = \frac{g(x)}{1/f(x)} \]
Either rewrite is algebraically correct. One of them will be far easier to differentiate than the other, and picking the wrong one is how a two-line problem becomes a page.
Intuition
Rule of thumb: move down whichever factor gets simpler when you take its reciprocal and differentiate.
Powers of the variable love being flipped - a reciprocal power is still just a power. Logarithms and inverse trig functions hate it: flipping them buries them inside a fraction and the chain rule makes them worse.
So the default is: leave the logarithm on top, send the power to the bottom.
Ranking
Put in order
Put the moves of Worked example: a product that fights itself into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The factor x approaches 0 and the natural log runs to negative infinity, so this is the zero times infinity form - indeterminate, but not yet a quotient.
Worked example
As the input shrinks toward zero from the right, the first factor dies and the second one runs to negative infinity.
\[ \lim_{x \to 0^{+}} x \ln x \]
Name the form
Why: The factor x approaches 0 and the natural log runs to negative infinity, so this is the zero times infinity form - indeterminate, but not yet a quotient.
Send the power downstairs, keep the logarithm on top
Why: The reciprocal of x is a clean negative power; the reciprocal of the logarithm would be a mess. This choice makes the derivatives easy.
\[ x \ln x = \frac{\ln x}{1/x} \]
Check the new form
Why: The numerator runs to negative infinity and the denominator grows without bound, so this is the infinity over infinity case and the rule now applies.
Apply the rule and simplify
Why: The derivative of the natural log is 1 over x. The derivative of the reciprocal is negative 1 over x squared. Dividing gives negative x, which is as simple as it gets.
\[ \lim_{x \to 0^{+}} \frac{1/x}{-1/x^{2}} = \lim_{x \to 0^{+}} (-x) = 0 \]
Verify with a table of shrinking inputs
Why: The product is heading to 0 from below. The logarithm blows up, but only logarithmically, and the factor x crushes it - the power wins the tug of war.
| x | value of the product |
|---|---|
| 0.1 | -0.230259 |
| 0.01 | -0.046052 |
| 0.001 | -0.006908 |
Picture it
Animation
Shows: Each line of the worked example "a product that fights itself", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The numerator runs to negative infinity and the denominator grows without bound, so this is the infinity over infinity case and the rule now applies.
Fill the middle
Fill in the blanks
From Trap: flipping the wrong factor — finish the line. Write what belongs on the right of the equals sign before you look.
x \ln x = \frac{x}{1/\ln x}
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The denominator's derivative requires the chain rule on a reciprocal logarithm, producing a compound fraction that is harder than the problem you started with.
Trap
Same limit, but the logarithm is the one sent downstairs.
\[ x \ln x = \frac{x}{1/\ln x} \]
The derivatives get worse, not better
Why: The denominator's derivative requires the chain rule on a reciprocal logarithm, producing a compound fraction that is harder than the problem you started with.
\[ \frac{1}{-\dfrac{1}{x(\ln x)^{2}}} = -x(\ln x)^{2} \]
You are now further from an answer
Why: The new expression is still the zero times infinity form, but with the logarithm squared. Each flip in this direction makes it worse.
Send the power down and keep the logarithm upstairs where its derivative is simple.
\[ x \ln x = \frac{\ln x}{x^{-1}} \;\longrightarrow\; \frac{1/x}{-x^{-2}} = -x \]
Check the outcome
Why: One application produces negative x, which visibly approaches 0. The correct choice of which factor to flip turned a hard problem into a one-liner.
Commit first
Predict first
What is the limit of x squared times the natural log of x as x approaches 0 from the right?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: Zero
Why: Rewrite as the natural log over x to the negative 2. One application gives 1 over x divided by negative 2 times x to the negative 3, which simplifies to negative x squared over 2, and that approaches 0. At x equal to 0.01 the product is about negative 0.00046.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
Convert it to a quotient first, then apply the rule.
\[ \lim_{x \to 0^{+}} x^{2} \ln x \]
Check your understanding
What is the limit of x squared times the natural log of x as x approaches 0 from the right?
Answer: A
Why: Rewrite as the natural log over x to the negative 2. One application gives 1 over x divided by negative 2 times x to the negative 3, which simplifies to negative x squared over 2, and that approaches 0. At x equal to 0.01 the product is about negative 0.00046.
Concept
A difference of two quantities that both grow without bound is a tug of war too. The size of the gap between them is exactly what is undecided.
Put both pieces over a common denominator. The subtraction disappears into a single fraction, and that fraction is almost always zero over zero.
If instead you are subtracting square roots, the common denominator is replaced by its cousin: multiply by the conjugate.
Hypothesis
Predict first
Worked example: a difference of two blow-ups is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Combine over a common denominator
Why: The common denominator is x times the quantity e to the x minus 1. This converts the indeterminate difference into a single quotient.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Each term separately runs to positive infinity as the input shrinks. The gap between them does something much calmer.
\[ \lim_{x \to 0^{+}} \left( \frac{1}{x} - \frac{1}{e^{x} - 1} \right) \]
Combine over a common denominator
Why: The common denominator is x times the quantity e to the x minus 1. This converts the indeterminate difference into a single quotient.
\[ \frac{e^{x} - 1 - x}{x\left(e^{x} - 1\right)} \]
Check the form
Why: Top at 0: 1 minus 1 minus 0 is 0. Bottom at 0: 0 times 0 is 0. Zero over zero, so the rule applies.
Differentiate top and bottom separately
Why: The bottom needs the product rule since it is x times a function, giving the exponential plus x times the exponential, minus 1.
\[ \lim_{x \to 0^{+}} \frac{e^{x} - 1}{e^{x} + x e^{x} - 1} \]
Re-check: still zero over zero
Why: Top at 0 is 0. Bottom at 0 is 1 plus 0 minus 1, which is 0. A second application is legal.
Apply the rule again
Why: The bottom's derivative is the exponential plus the product rule on x times the exponential, giving two copies of the exponential plus x times the exponential.
\[ \lim_{x \to 0^{+}} \frac{e^{x}}{2e^{x} + x e^{x}} = \frac{1}{2 + 0} = \frac{1}{2} \]
Verify with a table
Why: Both terms are enormous individually - at x equal to 0.001 they are about 1000 and 999.5 - yet their difference settles calmly on one half.
| x | value of the difference |
|---|---|
| 0.1 | 0.49167 |
| 0.01 | 0.49917 |
| 0.001 | 0.49992 |
Picture it
Animation
Shows: Each line of the worked example "a difference of two blow-ups", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Top at 0: 1 minus 1 minus 0 is 0. Bottom at 0: 0 times 0 is 0. Zero over zero, so the rule applies.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Both terms grow without bound, so surely they cancel.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The two terms do both grow without bound, but the square root stays a fixed distance ahead.
Multiply by the conjugate to turn the difference into a quotient.
Why: The two terms do both grow without bound, but the square root stays a fixed distance ahead. At x equal to 1000 the difference is 1.4989, not something shrinking toward 0.
Trap
Both terms grow without bound, so surely they cancel.
\[ \lim_{x \to \infty} \left( \sqrt{x^{2} + 3x} - x \right) \;\overset{?}{=}\; 0 \]
Infinity is not a number, so it cannot cancel
Why: The two terms do both grow without bound, but the square root stays a fixed distance ahead. At x equal to 1000 the difference is 1.4989, not something shrinking toward 0.
Multiply by the conjugate to turn the difference into a quotient.
\[ \sqrt{x^{2}+3x} - x = \frac{3x}{\sqrt{x^{2}+3x} + x} \]
Verify the value at a large input
Why: Dividing top and bottom by x gives 3 over the quantity root of 1 plus 3 over x, plus 1, which approaches 3 over 2. At x equal to 1000 the true difference is 1.4989, right beside 1.5.
Step zero
Discussion prompt
Worked example: the conjugate finishes it — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the form
Answer:
Worked example
The same limit, done carefully start to finish.
\[ \lim_{x \to \infty} \left( \sqrt{x^{2} + 3x} - x \right) \]
Name the form
Why: Both terms grow without bound, so this is the infinity minus infinity form - indeterminate and not yet a quotient.
Multiply by the conjugate over itself
Why: Multiplying by 1 in this disguised form clears the radical from the numerator, since the difference of squares kills the square root.
\[ \left(\sqrt{x^{2}+3x} - x\right)\cdot\frac{\sqrt{x^{2}+3x} + x}{\sqrt{x^{2}+3x} + x} = \frac{3x}{\sqrt{x^{2}+3x} + x} \]
Divide top and bottom by x
Why: The variable is heading to positive infinity, so x is positive and the square root of x squared is simply x - no absolute value complication in this direction.
\[ \frac{3}{\sqrt{1 + 3/x} + 1} \;\longrightarrow\; \frac{3}{1 + 1} = \frac{3}{2} \]
Verify numerically at a large input
Why: At x equal to 1000 the square root of 1003000 is about 1001.4989, so the difference is 1.4989 - agreeing with three halves. Note that L'Hopital's Rule on the conjugate form would have cycled here.
Concept
The last three indeterminate forms are all powers where the base and the exponent are both moving.
\[ 1^{\infty} \qquad 0^{0} \qquad \infty^{0} \]
All three get the same treatment, so you do not have to remember which is which. Name the whole expression, take its natural log, find that limit, then undo the log.
Picture it
Animation
Shows: The three exponential forms — a rendered Manim animation.
Rendered with Manim.
Takeaway: Two conversions before the rule can be used at all.
Intuition
A logarithm converts an exponent into a multiplier. That is the single property doing all the work here.
A power with two moving parts becomes a product with two moving parts, and you already know how to handle a product: send one factor downstairs and use the rule.
The catch is bookkeeping. The limit you compute after taking the log is the limit of the logarithm of your answer - not the answer. You still have to undo the log at the end.
Explain it
Discussion prompt
Explain Why a logarithm rescues a power to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A logarithm converts an exponent into a multiplier. That is the single property doing all the work here.
Pattern
For any of the three exponential forms, in this exact order.
Edge cases
Discussion prompt
The log-then-exponentiate recipe works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
For any of the three exponential forms, in this exact order.
Estimation
Predict first
The base creeps toward one while the exponent runs away.
Commit before you compute: what does Worked example: one to the infinity come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with large inputs
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. The values climb toward about 20.09, which is the natural exponential cubed - not 1, and not unbounded.
Worked example
The base creeps toward one while the exponent runs away.
\[ \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{x} \]
Name the form and the expression
Why: The base approaches 1 and the exponent grows without bound, so this is the one-to-the-infinity form. Call the expression y so we can take its logarithm.
Take the natural log and pull the exponent down
Why: The log of a power is the exponent times the log of the base, which converts the power into a product.
\[ \ln y = x \ln\!\left(1 + \frac{3}{x}\right) \]
Convert the product into a quotient
Why: The log factor approaches 0 and the x factor grows without bound, so this is zero times infinity. Send the x downstairs as its reciprocal.
\[ \ln y = \frac{\ln\!\left(1 + 3/x\right)}{1/x} \]
Apply the rule
Why: The numerator needs the chain rule: 1 over the base times the derivative of the base, which is negative 3 over x squared. The denominator's derivative is negative 1 over x squared, and those two negative squared terms cancel beautifully.
\[ \lim_{x \to \infty} \frac{-3/(x^{2}+3x)}{-1/x^{2}} = \lim_{x \to \infty} \frac{3x^{2}}{x^{2}+3x} = \lim_{x \to \infty} \frac{3x}{x+3} = 3 \]
Exponentiate to undo the logarithm
Why: What we found is the limit of the logarithm of y, not of y. Raising the natural exponential to that number recovers the limit we actually wanted.
\[ \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{x} = e^{3} \approx 20.086 \]
Verify with large inputs
Why: The values climb toward about 20.09, which is the natural exponential cubed - not 1, and not unbounded.
| x | value of the expression |
|---|---|
| 10 | 13.786 |
| 100 | 19.219 |
| 1000 | 19.996 |
| limit | 20.086 |
Picture it
Animation
Shows: Each line of the worked example "one to the infinity", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The values climb toward about 20.09, which is the natural exponential cubed - not 1, and not unbounded.
Fill the middle
Fill in the blanks
From Trap: stopping before you exponentiate — finish the line. Write what belongs on the right of the equals sign before you look.
\lim_x \ln x \;\longrightarrow\; 0} x^___: \quad \ln y = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The number 0 is the limit of the logarithm of the expression, not of the expression.
Trap
The logarithm limit is computed correctly, then reported as the answer.
\[ \lim_{x \to 0^{+}} x^{x}: \quad \ln y = x \ln x \;\longrightarrow\; 0 \]
Reporting 0 is wrong by the width of an exponential
Why: The number 0 is the limit of the logarithm of the expression, not of the expression. The values of the expression are near 1, not near 0 - at x equal to 0.001 the expression equals 0.9931.
Finish the job: undo the logarithm.
\[ \lim_{x \to 0^{+}} x^{x} = e^{0} = 1 \]
Check the size against a small input
Why: At x equal to 0.01 the expression is 0.9550 and at 0.001 it is 0.9931 - climbing toward 1, exactly as the exponentiated answer predicts. Write the phrase the limit of the log equals, so you never forget which quantity you found.
Notation
Annotate
From Trap: stopping before you exponentiate — read this one piece at a time. What is each part doing?
On: \( \lim_{x \to 0^{+}} x^{x} = e^{0} = 1 \)
Step zero
Discussion prompt
Worked example: zero to the zero — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Name the form
Answer:
Worked example
Base and exponent both shrink to zero at once.
\[ \lim_{x \to 0^{+}} x^{x} \]
Name the form
Why: The base approaches 0 and so does the exponent, giving the zero-to-the-zero form. A base heading to 0 pushes the value down; an exponent heading to 0 pushes it toward 1. Genuine tug of war.
Take the natural log
Why: The exponent comes down front and the power becomes a product we can handle.
\[ \ln y = x \ln x \]
Reuse the product limit we already proved
Why: We showed earlier that this product approaches 0 by rewriting it as the log over the reciprocal of x and applying the rule once.
\[ \lim_{x \to 0^{+}} x \ln x = 0 \quad\Longrightarrow\quad \lim_{x \to 0^{+}} \ln y = 0 \]
Exponentiate
Why: The natural exponential of 0 is 1, so the expression itself approaches 1 even though its base is collapsing.
\[ \lim_{x \to 0^{+}} x^{x} = e^{0} = 1 \]
Verify with a table of small inputs
Why: The values rise steadily toward 1 as the input shrinks, confirming that the exponent's pull toward 1 beats the base's collapse toward 0.
| x | value of the expression |
|---|---|
| 0.1 | 0.7943 |
| 0.01 | 0.9550 |
| 0.001 | 0.9931 |
Picture it
Animation
Shows: Each line of the worked example "zero to the zero", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The values rise steadily toward 1 as the input shrinks, confirming that the exponent's pull toward 1 beats the base's collapse toward 0.
Prediction
Predict first
What is the limit of the quantity one plus two over x, raised to the power x, as x grows without bound?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: The natural exponential squared, about 7.389
Why: Taking the natural log gives x times the log of one plus two over x, which converts to a quotient and yields 2. Exponentiating gives the natural exponential squared, about 7.389. At x equal to 1000 the expression is about 7.374.
Check
Take the log, find that limit, and remember the last step.
\[ \lim_{x \to \infty} \left(1 + \frac{2}{x}\right)^{x} \]
Check your understanding
What is the limit of the quantity one plus two over x, raised to the power x, as x grows without bound?
Answer: A
Why: Taking the natural log gives x times the log of one plus two over x, which converts to a quotient and yields 2. Exponentiating gives the natural exponential squared, about 7.389. At x equal to 1000 the expression is about 7.374.
Section
Part 5
Concept
Look again at the fine print in the statement: the two limits are equal provided the second one exists (or runs to infinity).
So the rule can hand you an answer, but it can never take one away. If the derivative ratio has no limit, you have learned nothing about the original - not that it fails, not that it succeeds.
This is the difference between a tool being inconclusive and a limit being nonexistent, and it is the most sophisticated point in this deck.
Analogy
Discussion prompt
Explain The rule is a one-way street by analogy to something with no Calculus I in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Look again at the fine print in the statement: the two limits are equal provided the second one exists (or runs to infinity).
Ranking
Put in order
Put the moves of Worked example: the rule refuses to answer into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The sine term is trapped between negative 1 and 1, so the numerator grows without bound along with x.
Worked example
A perfectly reasonable limit that L'Hopital's Rule cannot finish.
\[ \lim_{x \to \infty} \frac{x + \sin x}{x} \]
Name the form
Why: The sine term is trapped between negative 1 and 1, so the numerator grows without bound along with x. The form is infinity over infinity and the rule is legal.
Apply the rule and watch it fail
Why: The derivative of the numerator is 1 plus cosine x; the derivative of the denominator is 1. The result oscillates forever between 0 and 2 and has no limit.
\[ \lim_{x \to \infty} \frac{1 + \cos x}{1} = \lim_{x \to \infty} (1 + \cos x) \quad \text{does not exist} \]
Conclude nothing, and start over with algebra
Why: The hypothesis that the new limit exists has failed, so the rule is silent. Split the fraction instead.
\[ \frac{x + \sin x}{x} = 1 + \frac{\sin x}{x} \]
Squeeze the leftover piece
Why: The sine is bounded between negative 1 and 1 while the denominator grows without bound, so the second term is squeezed to 0 and the whole expression approaches 1.
\[ \left| \frac{\sin x}{x} \right| \le \frac{1}{x} \;\longrightarrow\; 0 \qquad \Longrightarrow \qquad \lim_{x \to \infty} \frac{x + \sin x}{x} = 1 \]
Verify at two large inputs
Why: At x equal to 100 the value is 0.99494 and at x equal to 1000 it is 1.00083 - hugging 1 from both sides, exactly as the squeeze predicts.
| x | value of the ratio |
|---|---|
| 100 | 0.99494 |
| 1000 | 1.00083 |
| limit | 1 |
Picture it
Animation
Shows: When the rule refuses to help — a rendered Manim animation.
Rendered with Manim.
Takeaway: If the new limit does not exist, the rule tells you nothing at all.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The derivative ratio oscillates, so the student writes that the original limit does not exist.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The rule only promises equality when the right-hand limit exists.
When the derivative ratio has no limit, discard the attempt and use another method.
Why: The rule only promises equality when the right-hand limit exists. Here it does not, so the chain of equalities breaks at exactly that link and carries no information.
Trap
The derivative ratio oscillates, so the student writes that the original limit does not exist.
\[ \lim_{x \to \infty} \frac{x + \sin x}{x} \;\overset{?}{=}\; \lim_{x \to \infty} (1 + \cos x) \;\overset{?}{=}\; \text{DNE} \]
That second equals sign was never earned
Why: The rule only promises equality when the right-hand limit exists. Here it does not, so the chain of equalities breaks at exactly that link and carries no information.
When the derivative ratio has no limit, discard the attempt and use another method.
\[ \lim_{x \to \infty} \left(1 + \frac{\sin x}{x}\right) = 1 + 0 = 1 \]
Check the claim numerically
Why: At x equal to 1000 the ratio is 1.00083, nowhere near oscillating between 0 and 2. The original limit exists and equals 1 - the rule simply could not see it.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
You can apply the rule to sine of x over x and get the right answer of 1. As a proof, though, it is circular.
\[ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1 \]
The step uses the derivative of sine - and the derivative of sine is established from the squeeze-theorem proof of this very limit. You would be assuming what you set out to show.
On a homework problem this is fine as a computation. If a question says prove, use the squeeze theorem instead.
Counterexample
Discussion prompt
You can apply the rule to sine of x over x and get the right answer of 1. As a proof, though, it is circular.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The step uses the derivative of sine - and the derivative of sine is established from the squeeze-theorem proof of this very limit. You would be assuming what you set out to show.
Concept
One dollar at five percent annual interest, compounded a certain number of times per year, is worth the base one plus the rate over the number of periods, raised to the number of periods.
Compound it more and more often and you get exactly the one-to-the-infinity form. Its limit is what banks call continuous compounding.
\[ \lim_{n \to \infty} \left(1 + \frac{r}{n}\right)^{n} = e^{r} \]
| Compounding periods per year | Value of one dollar after a year |
|---|---|
| 1 | 1.050000 |
| 12 | 1.051162 |
| 365 | 1.051267 |
| continuous | 1.051271 |
The gain flattens out fast. That flattening is the indeterminate form resolving itself, and it is worth about one tenth of a cent per dollar.
Pattern
Step through it
Step through Where one-to-the-infinity shows up for real one row at a time. What is driving the change, and what would the row after the last one be?
Concept
L'Hopital's Rule is one tool among several. These are the situations where something else is faster, safer, or the only option.
| Situation | Better tool |
|---|---|
| Ratio of polynomials as the input grows | divide by the highest power in the denominator |
| Zero over zero in a rational function at a point | factor and cancel |
| A difference or sum involving a square root | multiply by the conjugate |
| A bounded oscillating piece over something growing | bound it and squeeze |
| The derivative ratio cycles back to the start | algebra - the rule will never terminate |
| The form is not indeterminate at all | just substitute |
Comparison
Comparison matrix
From What to reach for when the rule is wrong for the job: refill the Better tool column from what you know. The rest of the table is as it appeared.
| Situation | Better tool |
|---|---|
| Ratio of polynomials as the input grows | divide by the highest power in the denominator |
| Zero over zero in a rational function at a point | factor and cancel |
| A difference or sum involving a square root | multiply by the conjugate |
| A bounded oscillating piece over something growing | bound it and squeeze |
| The derivative ratio cycles back to the start | algebra - the rule will never terminate |
| The form is not indeterminate at all | just substitute |
Pattern
This is the flowchart to run on every limit you meet from here on.
Real world
Discussion prompt
Outside this lesson: where does Indeterminate Forms and L'Hopital's Rule actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of The complete decision procedure is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck explains what makes a limit form indeterminate rather than merely undefined, lists the seven indeterminate forms, and applies L'Hopital's Rule with its hypotheses checked every single time. It targets the four classic errors: applying the rule to a determinate form, using the quotient rule instead of differentiating the top and the bottom separately, skipping the re-check before a second application, and forgetting to exponentiate at the end of a log-then-limit problem.
Elimination
Eliminate the wrong options
What is the limit of the square root of the quantity x squared plus 5x, minus x, as x grows without bound?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Multiplying by the conjugate gives 5x over the square root of x squared plus 5x, plus x. Dividing top and bottom by x gives 5 over the quantity approaching 2, which is five halves. At x equal to 1000 the true difference is 2.4969.
Check
Name the form first, then choose your tool.
\[ \lim_{x \to \infty} \left( \sqrt{x^{2} + 5x} - x \right) \]
Check your understanding
What is the limit of the square root of the quantity x squared plus 5x, minus x, as x grows without bound?
Answer: A
Why: Multiplying by the conjugate gives 5x over the square root of x squared plus 5x, plus x. Dividing top and bottom by x gives 5 over the quantity approaching 2, which is five halves. At x equal to 1000 the true difference is 2.4969.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — What Indeterminate Really Means · L'Hopital's Rule · Infinity over Infinity and Growth Rates · Converting the Other Five Forms · When the Rule Says Nothing. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can tell an indeterminate form from a determinate one, and you know that the form is a signal to work harder - never an answer.
| Form | Your move |
|---|---|
| zero over zero | apply the rule directly |
| infinity over infinity | apply the rule directly |
| zero times infinity | send one factor down as a reciprocal |
| infinity minus infinity | common denominator, or conjugate |
| one to the infinity, zero to the zero, infinity to the zero | natural log, then the limit, then exponentiate |
And you know the two silences: the rule says nothing when the form was never indeterminate, and nothing when the derivative ratio has no limit. In both cases, algebra is waiting.
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