If $f \in C [0,1],\lim _{x \to 0+} f(x)/x$ exists finitely, then$\lim_{n \to \infty} n\left(n \int_0^1 f(x^n)dx-\int_0^1 \frac {f(x)}x dx\right)=$?

Question

Let $f:[0,1] \to \mathbb R$ be a continuous function such that $\lim\limits _{x \to 0+} \frac {f(x)}x$ exists finitely . Then does the limit $\lim\limits_{n \to \infty} n\left(n \int_0^1 f(x^n)dx-\int_0^1 \frac {f(x)}x dx\right)$ exist ? If it does then what is the value of the limit ? Please help . Thanks in advance

Answer 1

Hint:

The given problem can be rewritten in the form

\begin{align} \lim\limits_{n\rightarrow \infty}n\left(\int\limits_{0}^{1} \frac{f(x)}{x}\left(x^{1/n}-1\right)\right)&=\lim\limits_{n\rightarrow \infty}n\left(\int\limits_{0}^{1} \frac{f^{+}(x)}{x}\left(x^{1/n}-1\right)\right)-\lim\limits_{n\rightarrow \infty}n\left(\int\limits_{0}^{1} \frac{f^{-}(x)}{x}\left(x^{1/n}-1\right)\right). \end{align}

Can you proceed by checking the behaviour of $g_{n}(x)=n\left(x^{1/n}-1\right)$, $x\in (0,1]$, with respect to $n$? How do you justify pushing the limit under the integral sign in each of the terms on the RHS?