If $g:[0,1] \rightarrow \mathbb{R}$ be a continuous function such that $\lim\limits_{x \rightarrow 0^+}\dfrac{g(x)}{x}$ exist and is finite, then prove that
$$\lim\limits_{n \rightarrow \infty} \int_{0}^{1}g(x^n)dx= \int_{0}^{1} \frac{g(x)}{x}dx$$
This is problem from the book PROBLEMS IN REAL ANALYSIS- ADVANCE CALCULUS ON THE REAL AXIS by Titu Andreescu. The solution given on the book can't be understood by me. Please provide any alternative solution.
On
This is false. Consider $g(x) = x$, then $$\int_0^1 g(x^n)dx = \int_0^1 x^ndx = \frac{x^{n+1}}{n+1}|_{x=0}^{x=1} = \frac{1}{n+1} \rightarrow 0$$ $$\int_0^1 \frac{g(x)}{x}dx = \int_0^1 dx = 1$$
Edit: It is true however that $$\lim_{n\rightarrow \infty} \int_0^1 n g(x^n)dx = \int_0^1 \frac{g(x)}{x}dx$$
If you change variables you find $$\int_0^1 n g(x^n)dx = \int_0^1 y^{1/n} \frac{g(y)}{y}dy$$ and apply dominated convergence as $$y^{1/n} \frac{g(y)}{y} \rightarrow \frac{g(y)}{y} \text{ for } y\in(0,1]$$ which is true from our assumption on $g$.
Define $h:[0,1] \rightarrow \mathbb{R}$ by
$h(t) = \begin {cases} g(t)/t , & t \in (0,1] \\ lim_{x \rightarrow 0^+}~g(x)/x, & t = 0 \end{cases} $
Then $h$ is continuous and we can set $H(x) = \int_0^1 h(t)dt.$
We have
$n \int_0^1g(x^n)dx=n \int_0^1x^nh(x^n)dx=xH(x^n)|_0^1- \int_0^1H(x^n)dx=H(1) - \int_0^1H(x^n)dx= \int_0^1 \frac{g(x)}{x}dx - \int_0^1H(x^n)dx.$
If $0 < a<1$, then
$| \int_0^1H(x^n)dx| \leq \int_0^1|H(x^n)|dx = \int_0^a|H(x^n)|dx+ \int_a^1|H(x^n)|dx \leq a|H( \alpha _n^n)| + (1-a)M~~~~~~~(1)$
where $\alpha _n^n \in [0,a]$ and $M=max_{t \in [0,1]}|H(t)|.$
Consider $ \epsilon > 0$ such that $a>1 - \frac{\epsilon}{2M}$. Since $lim_{n \rightarrow \infty}|H( \alpha _n^n)|=0$ it follows that $a|H( \alpha _n^n)|< \epsilon /2$ for all positive itegers $n \leq N(\epsilon)$. Relation (1) yields
$| \int_0^1H(x^n)dx| \leq \epsilon /2 + (1-a)M < \epsilon /2 + (1-1+ \epsilon /2M)M= \epsilon$
Hence, $lim_{n \rightarrow \infty} \int_0^1 H(x^n)dx= 0$ and the conclusion follows.