Proof

Let $D\subseteq\mathbb R$ be a subset of real numbers.
Let $f,g,h:D\to\mathbb R$ be functions with $g(x)\le f(x)\le h(x)$ for all $x\in D$.
Let $g,h$ have a limit $L$ at $x=a\in D,$ i.e. $\lim_{x\to a} g(x)=\lim_{x\to a}h(x)=L.$
Let $\epsilon > 0$ be given.
By the definition of limit, there is a $\delta_1 > 0$ such that for all $x\in D$ with $0 < |x-a| < \delta_1,$ it follows $|g(x)-L| < \epsilon.$
Similarly, there is a $\delta_2 > 0$ such that for all $x\in D$ with $0 < |x-a| < \delta_2,$ it follows $|h(x)-L| < \epsilon.$
Take the minimum $\delta:=\min(\delta_1,\delta_2).$
For $x\in D$ with $0 < |x-a| < \delta,$ it follows $|g(x)-L| < \epsilon$ and $|h(x)-L| < \epsilon.$
Thus, for those $x,$ we have that $-\epsilon < g(x)-L\le f(x)-L\le h(x)-L < \epsilon.$
For those $x,$ it follows that $|f(x)-L| < \epsilon.$
By the definition of limit, $\lim_{x\to a} f(x)=L.$
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