Is a function differentiable in $x$ if $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\infty$?

Question

All the definitions of differentiability I found (Wolfram Mathworld for instance) only require this limit to exist, but say nothing about the domain in which that has to happen.

So what if that limit is $\pm\infty$? Wouldn't (in the "normal" interpretation of a function $\mathbb R \rightarrow \mathbb R$, for instance) the slope of the tangent of $f$ in $x$ be undefined?

Edit: a link clearing it up.

Answer 1

This is an unfortunate case of terminology getting the better of us. The definition of a limit is

The limit of the function $f(x)$ as $x$ approaches $a$ exists if there is a number $L$ such that for all $\epsilon>0$, there exists a $\delta>0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$. In this case, we say that $$\lim_{x\to a} f(x)=L.$$

So the limit, if it exists is a number. Infinity is not a number, so technically, if $\lim_{x\to a}f(x)=\infty$, the limit doesn't exist. And yet it is infinity. It's confusing at first, but once you get used to it, it's not too bad.

Answer 2

For a function $f: \mathbb{R} \to \mathbb{R}$ to be differentiable, we require that $$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = L,$$ where $L \in \mathbb{R}$. Sometimes the wording is "the limit exists and is finite," but other times the finite portion is dropped, and the phrase "limit exists" means, in particular, that the limit is not $\infty$.