Sandwich (Squeeze) Theorem Illustrator

Example & Functions

g(x) = −x²

f(x) = x²·sin(1/x)

h(x) = x²

x → a = 0

g(x) =

f(x) =

h(x) =

a =

Showing the classic squeeze: −x² ≤ x²sin(1/x) ≤ x², all tending to 0 as x → 0.

Limit Result & Verification

Concluded by Sandwich Theorem

lim_x→0 x²sin(1/x)

= 0

Numerical Verification: g(x) ≤ f(x) ≤ h(x) near x = a

x	g(x)	f(x)	h(x)	g≤f≤h?

The Sandwich Theorem (Formal Statement)

If g(x) ≤ f(x) ≤ h(x) near x = a,
and lim g(x) = lim h(x) = L,
then lim f(x) = L.

Three conditions must all hold:

Bounding inequality — g(x) ≤ f(x) ≤ h(x) must hold for all x sufficiently close to a (but not necessarily at a itself).
Common limit from below — lim_x→a g(x) = L must exist and equal L.
Common limit from above — lim_x→a h(x) = L must also equal L.

When all three hold, f is forced to converge to L — it is squeezed between two functions that both converge to the same value.

Note: f(x) does not need to be defined at x = a for the limit to exist. This is exactly what makes it powerful for functions like x²sin(1/x).

Why the Sandwich Theorem Is Useful

f(x) = x²·sin(1/x) — direct limit fails at x=0

Some functions are hard or impossible to evaluate directly at a limit point:

x²sin(1/x) — sin(1/x) oscillates wildly near 0, so substitution fails. But |sin(1/x)| ≤ 1 gives us −x² ≤ x²sin(1/x) ≤ x², and both bounds → 0.
sin(x)/x — classic indeterminate form 0/0 at x = 0. The geometric inequality cos(x) ≤ sin(x)/x ≤ 1 near 0 squeezes the limit to 1 — the foundation of all trig derivative formulas.
General strategy — use a known bound (often |f(x)| ≤ M·|x−a|ⁿ for some M and n > 0), then both bounding functions tend to 0.

The Sandwich Theorem also appears in real analysis (proving convergence of sequences) and multivariable calculus (showing 2-variable limits exist).