Solutions

Suppose that $f(x) \le g(x) \le h(x)$ and that

$$\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = l.$$

By definition, there exist $\delta_1$ and $\delta_2$ such that

$$0<|x-a|<\delta_1 \implies |f(x)-l|<\epsilon,$$$$0<|x-a|<\delta_2 \implies |h(x)-l|<\epsilon.$$

Let $\delta = \min(\delta_1,\delta_2)$. We have

$$0<|x-a|<\delta \implies -\epsilon+l < f(x) < \epsilon+l,$$$$-\epsilon+l < h(x) < \epsilon+l.$$

But $f(x) \le g(x) \le h(x)$ must also apply when $0<|x-a|<\delta$. Hence,

$$-\epsilon+l < f(x) \le g(x) \le h(x) < \epsilon+l$$

holds, and $\lim_{x \to a} g(x)$ exists and is $l$.