Theorem: Let $b< a < c$ be real numbers, suppose $f(x)\le g(x)\le h(x)$ for all $x\in(b,a)\cup(a,c)$ and $\lim_{x\to a}\ f(x)=L=\lim_{x\to a}\ h(x)$. Then $\lim_{x\to a}\ g(x)=L$.
Proof:

Suppose the assumptions above are satisfied, and consider any $\epsilon >0$.

To show that $\lim_{x\to a}\ g(x)=L$, we must find a $\delta>0$ such that if $0< |x-a|< \delta$ then $|g(x)-L|< \epsilon$.

Since $\lim_{x\to a}\ f(x)=L$ we can find a $\delta_1>0$ such that if $0< |x-a|< \delta_1$ then $|f(x)-L|< \epsilon$.

Since $\lim_{x\to a}\ h(x)=L$ we can find a $\delta_2>0$ such that if $0< |x-a|< \delta_2$ then $|h(x)-L|< \epsilon$.

Note that the conclusions in the previous two sentences are equivalent to $L-\epsilon< f(x)< L+\epsilon$ and $L-\epsilon< h(x)< L+\epsilon$.

Take $\delta=\text{min}\{\delta_1,\delta_2\}$ and assume $0< |x-a|< \delta$.

Then $0< |x-a|< \delta_1$ and $0< |x-a|< \delta_2$, so $L-\epsilon< f(x)\le g(x)\le h(x)< L+\epsilon$.

Hence we have shown that if $0< |x-a|< \delta$ then $|g(x)-L|< \epsilon$.

This proves that $\lim_{x\to a}\ g(x)=L$.