Limit is plus infinity, limit of square root is?

Question

Suppose for a continuous function $f$, with $ x_{0} \in D_{f}$, that $\lim_{x \to x_{0}}f(x) = +\infty$. Prove that $\lim_{x \to x_{0}}\sqrt {f(x)} = +\infty$.

Any ideas? I know this is very intuitive but can anyone post a basic proof?

Answer 1

Let $K>0$. You need to find $\delta>0$ such that, for $0<|x-x_0|<\delta$, $\sqrt{f(x)}>K$, which is the same as saying $f(x)>K^2$. Apply the definition for $\lim_{x\to x_0}f(x)=\infty$.