Squeeze Theorem

Theorem: Let $$p$$ be a point on interval $$I$$. Also let functions $$f$$, $$g$$ and $$h$$ be defined on the interval, except for possibly at point $$p$$.

Suppose that, $$\forall x\neq a\in{I}$$ that $$g(x)\leq f(x)\leq h(x)$$

and that $$\lim_{x\rightarrow a}g(x)=\lim_{x\rightarrow a}h(x)=L$$

Then, $$\lim_{x\rightarrow a}f(x)=L$$

Proof: We start by proving the special case where $$g(x)=0\mbox{ }\forall x\mbox{ and }L=0$$, in this case $$\lim_{x\to a}h(x)=0$$ Let $$\epsilon$$ be a positive integer, then by the definition of the limit of a function there is $$\delta>0$$ such that $$\mbox{if }0<|x-a|<\delta,\mbox{ then }|h(x)|<\epsilon$$ For any $$x\neq a,\mbox{ }0=g(x)\leq f(x)\leq h(x)$$ so that $$|f(x)|\leq|h(x)|$$ From this we conclude that $$\mbox{if }0<|x-a|<\delta,\mbox{ then }|f(x)|\leq|h(x)|<\epsilon$$ By the transitive property, this proves that $$\lim_{x\to a}f(x)=0=L$$ We now move on to the general case, with $$g(x)\mbox{ and }L$$ arbitrary. For $$x\neq a$$, we have $$g(x)\leq f(x)\leq h(x)$$ By subtracting $$g(x)$$ from all expressions, we have $$0\leq f(x)-g(x)\leq h(x)-g(x)$$ Since as $$x\rightarrow a, h(x)\rightarrow L\mbox{ and } g(x)\rightarrow L$$, we have $$h(x)-g(x)\rightarrow L-L=0$$ From the special case, we now have $$f(x)-g(x)\rightarrow 0$$ We conclude that $$f(x)=(f(x)-g(x))+g(x)\rightarrow 0+L=L$$ Q.E.D.