Squeeze Theorem

Theorem 4.1   If $${\lim_{x\to a} f(x)=L}$$ and $${\lim_{x\to a} h(x)=L}$$ and $f(x)\leq g(x) \leq h(x)$ then $${\lim_{x\to a} g(x)=L}$$.

PROOF:

Given $\epsilon>0$ we can use $${\lim_{x\to a} f(x)=L}$$ to get $\delta_f$ such that if $0<\vert x-a\vert<\delta_f$ then $L-\epsilon<f(x)<L+\epsilon$. We can use $${\lim_{x\to a} h(x)=L}$$ to get $\delta_h$ such that if $0<\vert x-a\vert<\delta_h$ then $L-\epsilon<h(x)<L+\epsilon$. Let $\delta=\min(\delta_f,\delta_h)$ then if $0<\vert x-a\vert<\delta$ then

$$L-\epsilon < f(x) \leq g(x) \leq h(x) < L+\epsilon.$$

This tells us that $\vert g(x)-L\vert<\epsilon$.