# Squeeze theorem with infinite limits

Let f,g be functions that are defined in the area of $x_0$ (Except $x_0$ itself)

$f(x) \ge g(x)$

Given the limit $\lim_{x \to x_0}g(x) = \infty$

Prove that $\lim_{x \to x_0}f(x) = \infty$

It seems logic that if $g(x)$ approaches some value where its height (limit) is at infinity, any other functions above $g(x)$ are at infinity too in this area, but I don't know what is the right approach proving this.

some help? :)

Definition of $\lim_{x \to x_0} g(x) = \infty$ is that for all $M \in \Bbb{R}$ there exists $\delta > 0$ such that $|x-x_0| < \delta$ implies $g(x) > M$. Now, let $M \in \Bbb{R}$ be arbitrary. Because $\lim_{x \to x_0} g(x) = \infty$, we know that there exists $\delta$ such that \begin{align*} |x-x_0| < \delta \implies f(x) \geq g(x) > M \end{align*} So $\lim_{x \to x_0} f(x) = \infty$.
You can always think to this: $$\lim_{x\to x_0}f(x)=\infty \qquad\text{if and only if}\qquad \lim_{x\to x_0} \arctan(f(x))=\frac{\pi}{2}$$ so you can apply the “finite squeeze theorem”.
Similarly, $$\lim_{x\to \infty}f(x)=l \qquad\text{if and only if}\qquad \lim_{t\to \frac{\pi}{2}^-} f(\tan t)=l$$ and you can combine the two for infinite limits at infinity (or for $-\infty$).