# Is it true that $\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\pm\infty\ \implies\ \lim_{x \to a}f'(x) = \pm\infty$?

Is it true that $$\lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\pm\infty \implies \lim_{x \to a}f'(x) = \pm\infty$$?

Here, $$f$$ is a function defined on some open interval $$I$$, and $$a\in I$$. Assume $$f$$ is continuous at $$a$$ and differentiable around $$a$$.

I can't for the life of me see how to prove\disprove this implication, but my gut feeling is that it's false. Any guidance is greatly appreciated.

• isn't that definition of limit? the (ε, δ) should provide that the function is strictly increasing and must go unbounded to $\infty$ – user29418 Jan 7 at 1:28
• Derivative of a differential function need not be continuous – Sorfosh Jan 7 at 6:00

The claim is false. Consider $$g(x)=\sqrt{x}\sin\frac{1}{x}+x^{1/4},\qquad x>0,$$ and define $$f$$ on $$\mathbb R$$ by $$f(x)=\begin{cases}g(x) & \text{for }x>0, \\ 0 & \text{for }x=0, \\ -g(-x) & \text{for }x<0.\end{cases}$$ Then $$f$$ is continuous everywhere, differentiable in $$\mathbb R\backslash\{0\}$$ and satisfies $$f(-x)=-f(x)$$. Moreover, for $$x>0$$, \begin{align*} \frac{f(x)}{x}=\frac{1}{\sqrt{x}}\sin\frac{1}{x}+\frac{1}{x^{3/4}}\geq-\frac{1}{x^{1/2}}+\frac{1}{x^{3/4}}\longrightarrow+\infty,\quad\text{as }x\to0+. \end{align*} Due to the symmetry, the same is true for $$x<0$$ and $$x\to0-$$. Thus, $$f(x)/x\to+\infty$$ as $$x\to0$$.
Now, for $$x>0$$, $$f'(x)=\frac{x^{3/4}+2 x \sin \left(\frac{1}{x}\right)-4 \cos \left(\frac{1}{x}\right)}{4 x^{3/2}}.$$ However, the limit $$\lim_{x\to0+}f'(x)$$ does not even exist. For $$x_n:=1/(n\pi)$$ we have \begin{align*} f'(x_n)=\frac{\pi ^{3/4}}{4 \left(\frac{1}{n}\right)^{3/4}}-\frac{\pi ^{3/2} (-1)^n}{\left(\frac{1}{n}\right)^{3/2}}, \end{align*} and so $$f'(x_{2n})\to-\infty$$, while $$f'(x_{2n+1})\to+\infty$$.