TITLE: Show that $\lim_{x\to 0^+} xf'(x)=0$.
QUESTION [5 upvotes]: Suppose $f$ is a continuous function on $[0,1]$. Suppose $f$ is differentiable on $(0,1)$ and its derivative is continuous on $(0,1)$. Then is it true that
$$\lim_{x\to 0^{+}} xf'(x)=0 \ ?$$
I only thought about functions like $x^{a}$ for $a>0$. It seems to be true. Indeed, in fact if we assume $f'$ is monotonically decreasing and non negative then $$0\le xf'(x) \le \int_0^x f'(t) \, dt = f(x)-f(0)$$
But I am not sure how to do it generally. Any help suggestions?

REPLY [6 votes]: Nope. Consider the function $$f(x) =\begin{cases}
x\sin(\frac{1}{x})\hspace{4mm}x > 0 \\
0 \hspace{17mm}x = 0\end{cases}$$
Then on $(0,1)$ we have $f'(x) = \sin(\frac{1}{x}) -\frac{1}{x}\cos(\frac{1}{x})$. Hence $$ \lim_{x\to0^+}xf'(x) = \lim_{x\to 0^+}x\sin\Big(\frac{1}{x}\Big)-\cos\Big(\frac{1}{x}\Big) = DNE$$
So the limit does not even exist.