TheoremProved

Squeeze theorem

Statement

If $g(x)\le f(x)\le h(x)$ for all $x$ near $a$ (except possibly at $a$ ) and $\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L$ , then $\lim_{x\to a} f(x) = L$ .

Why is it true?

A quantity trapped between two others that both converge to the same value has no room left to go anywhere else - like a person squeezed between two walls that both close in on the same spot.

Proof sketch

Fix $\varepsilon>0$ . Since $g(x)\to L$ and $h(x)\to L$ , there is $\delta>0$ such that for $0<|x-a|<\delta$ , both $|g(x)-L|<\varepsilon$ and $|h(x)-L|<\varepsilon$ , i.e. $L-\varepsilon<g(x)$ and $h(x)<L+\varepsilon$ . Combined with $g(x)\le f(x)\le h(x)$ , this gives $L-\varepsilon<f(x)<L+\varepsilon$ , i.e. $|f(x)-L|<\varepsilon$ , for all such $x$ . Since $\varepsilon$ was arbitrary, $f(x)\to L$ .

Topics that use this theorem

Limits of functions

Related theorems

Intermediate value theorem Bolzano–Weierstrass theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

James Stewart (2015). Calculus
Walter Rudin (1976). Principles of Mathematical Analysis