If $f : [0, 1] \to \mathbb{R}$ is continuous and differentiable on $(0, 1]$ and $\lim_{x \to 0} f'(x) = \infty$, then f is differentiable at 0

Question

Intuitively, it seems that $f$ can't even be continuous at 0, since the derivative is exploding up as $x$ tends to 0. Thus, $f$ is not differentiable at 0. I tried to prove that $f$ is not continuous at 0 using this fact but have been unable to make progress, as I'm having trouble relating the values of $f$ and the values of $f'$

Answer 1

With the given hypotheses, it's always the case that $f'(0)$ fails to exist as a finite number. In fact for each small $x>0,$ the MVT says there exists $c_x\in (0,x)$ such that

$$\frac{f(x)-f(0)}{x-0} = f'(c_x).$$

As $x\to 0^+,$ $c_x$ is dragged along to $0.$ By our hypothesis on $f'$ it follows that $f'(c_x) \to \infty,$ which implies $f'(0)=\infty$ (from the right) by the definition of the derivative.

Answer 2

Your condition $\lim_{x \to 0} f'(x) = \infty$ means that the tangent lines around zero tend to become vertical as $x$ approaches 0.

So any function differentiable near 0 but also continuous at 0 with a vertical tangent line to the graph at 0 satisfies your condition.

Automatically this means that the function is not differentiable at 0.

Answer 3

Just some notes about the question. With $f(x)=x^{2}\cos(1/x^{2})$, $x\ne 0$, $f(0)=0$, then $f$ exists on $[0,1]$ with $f'(0)=0$, $f'(x)=2x\cos(1/x^{2})+(1/x)\sin(1/x^{2})$, $x\in(0,1]$.

We see that there is a sequence $(x_{n})$ in $(0,1]$ such that $f'(x_{n})\rightarrow\infty$, but this does not mean that $\lim_{x\rightarrow 0}f'(x)=\infty$.