TITLE: Multiple choice exercise on $f(x)= \frac {\sin x}{|x|+ \cos x}$
QUESTION [5 upvotes]: Let $f : \Bbb R \to \Bbb R$ be the function defined by $f(x)= \frac {\sin x}{|x|+ \cos x}$. Then
A.$f$ is differentiable at all $x \in \Bbb R$.
B.$f$ is not differentiable at $x =0$.
C.$f$ is differentiable at $x=0$ but $f'$ is not continuous at $x=0$.
D.$f$ is not differentiable at $x=\frac {\pi}{2}$.

Using the standard definition of derivative, I get that $f$ is differentiable at $0$ as well as $\frac {\pi}2$. This is as follows,
$\lim_{h \to 0} \frac {f(0+h)-f(0)}{h}=1=\lim_{h \to 0} \frac {f(0-h)-f(0)}{-h}$
and
$\lim_{h\to 0} \frac {f(\frac {\pi}2 +h)-f(\frac {\pi}{2})}{h}=0=\lim_{h \to 0} \frac {f(\frac {\pi}2 -h)-f(\frac {\pi}2)}{-h}$ using L'Hospital rule at one stage.
Hence I eliminate options B and D.
Since $0$ was the only doubtful point to check about differentiability, I choose option A as the answer.
I regret a little about not knowing how to check validity of option C. Can you tell me how to prove it wrong?

REPLY [3 votes]: As per the discussion with @Quintic in the above comments,
$f'(x)=\frac {1+|x|\cos x-\frac {x}{|x|} \sin x}{(|x|+\cos x)^2}$.
Now, $\lim_{h \to 0} f'(0+h)=\lim_{h \to 0} \frac {1+|h|\cos h-\frac {h}{|h|}\sin h}{(|h|+\cos h)^2}=\lim_{h \to 0}\frac {1+h\cos h-\sin h}{(h+\cos h)^2}$. (Where $h \gt 0$)
$\Rightarrow \lim_{h\to 0} f'(0+h)=\frac {1+0-0}{(0+1)^2}=1.$
Similarly, $\lim_{h\to 0} f'(0-h)=\lim_{h\to 0} \frac {1+h\cos h-\sin h}{(h+\cos h)^2}$ (Where $h \gt 0)$
$\Rightarrow \lim_{h\to 0} f'(0-h)=1$.
Hence $\lim_{h\to 0} f'(0+h)=\lim_{h\to 0} f'(0-h) \Rightarrow$ $f'$ is continuous at $x=0$.