Show that $\lim_{x\to 0^+} xf'(x)=0$.

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.

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Limits

Integration

Derivatives

Real Analysis

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