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? 🙂
Best Answer
Definition of $\lim_{x \to x_0} g(x) = \infty$ is that for all $M \in \Bbb{R}$ there exists $\delta > 0$Â such that $|x-x_0| < \delta$ implies $g(x) > M$. Now, let $M \in \Bbb{R}$Â 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 $\lim_{x \to x_0} f(x) = \infty$.