Theorem. Let $I$ be an interval having the point $a$ as a limit point. Let $g$, $f_1$, $f_2$ ,..., $f_n$ and $h$ be functions defined on $I$, except possibly at $a$ itself. Given that for every $x$ in $I$ not equal to $a$, we have $$g(x)\leq f_1(x)\leq f_2(x)\leq ...\leq f_n(x)\leq h(x)$$ where $n\in\mathbb{Z^+}$ and that $\lim_{x\to a}g(x) = \lim_{x\to a}h(x) = N$, it is then that $$\lim_{x\to a}f_1(x)=\lim_{x\to a}f_2(x)=...=\lim_{x\to a}f_n(x)=N$$ Proof. Assume that $g(x)\leq f_1(x)\leq f_2(x)\leq ...\leq f_n(x)\leq h(x)$ and that $\lim_{x\to a}g(x)=\lim_{x\to a}h(x)$, by definition we may conclude the following $$\forall\epsilon_1>0\exists \delta_1>0\forall x(0<|x-a|<\delta_1\implies|g(x)-N|<\epsilon_1)\ \ (1)$$ $$\forall\epsilon_2>0\exists \delta_2>0\forall x(0<|x-a|<\delta_2\implies|h(x)-N|<\epsilon_2)\ \ (2)$$ let $k$ be arbitrary and assume that $k\in\{1,2,...,n\}$ consider now for $f_k$ an arbitrary $\epsilon>0$ and let $\delta = \textit{min}(\delta_1,\delta_2)$, let $x$ be arbitrary and assume that $0<|x-a|<\delta$, from $(1)$ and $(2)$ it follows that $|g(x)-N|<\epsilon$ and $|h(x)-N|<\epsilon$, but since $g(x)\leq f_k(x)\leq h(x)$ it is evident that $N-\epsilon<g(x)<f_k(x)<h(x)< N+\epsilon$ thus $|f(x)-N|<\epsilon$. \We may now conclude that $$\forall\epsilon>0\exists \delta>0\forall x(0<|x-a|<\delta\implies|f_k(x)-N|<\epsilon)\ \ \ (3)$$ consequently $\lim_{x\to a}f_k(x)=N$, as $k$ was arbitrary we may now generalize the above argument and conclude that $$\forall k\in\{1,2,...,n\}(\lim_{x\to a}f_k(x)= N)$$
$\blacksquare$
Is the above proof correct?