Limit of infinity times 0

Question

I have a question regarding a specific step in the proof of the theorem that 'differentiability implies continuity'.

The proof in my calculus book asserts that if $h\to0$ then:

$$\frac{f(x+h)-f(x)}{h}=f'(x)$$ Therefore: $$f(x+h)-f(x)=f'(x)h=f'(x)0=0$$

I can imagine cases where $f'(x)$ may be infinite, such as in the case of $g'(0)$ when $g(x)=x^\frac{1}{3}$

Question: how can one be sure that $f'(x)0$ is always $0$, even in the case of an infinite slope?

Answer 1

The key point ultimately boils down to:

If $\lim_{n\to\infty} a_n$ and $\lim_{n\to\infty} b_n$ exist, then $\lim_{n\to\infty} a_nb_n$ exists and equals $\left(\lim_{n\to\infty} a_n\right)\left(\lim_{n\to\infty} b_n\right)$.

Recall that existence of a limit (within $\Bbb R$) means specifically that the limit is a real number, so finite.

Answer 2

Note that the hypothesis of the theorem is differentiability; hence the theorem is not applicable to non-differentiable functions. A function $f$ not differentiable at a point $c$ is by definition such that $\lim_{h \to 0}[f(c+h) - f(c)] /h$ does not exist. The function $x \mapsto x^{1/3}$ is, for example, not differentiable at $0$; for we have $$ h^{1/3}/h = h^{-2/3} \to \infty $$ as $h \to 0.$