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. 
 A: 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 $f'(x_{2n+1})\to+\infty$.
