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.

  • $\begingroup$ Please include the hypothesis that $g$ is continuous in the statement of the problem (putting it vaguely in the title isn't good enough) ... $\endgroup$ Sep 11, 2019 at 17:01

2 Answers 2


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 $g$.

  • $\begingroup$ Yes You are right..But the strange thing is this is proved in the book..and this is a reputed book..my goodness.. Could we trust anyone..it is really hurting.. $\endgroup$ Sep 11, 2019 at 16:43
  • $\begingroup$ Can you provide the proof? It's possibly a typo on the question. $\endgroup$
    – Dayton
    Sep 11, 2019 at 16:44
  • $\begingroup$ Yes I can. Yeah I also think that there is a typo in the question. $\endgroup$ Sep 11, 2019 at 16:54
  • $\begingroup$ Perhaps my edit fixes the mistake? $\endgroup$
    – Dayton
    Sep 11, 2019 at 16:59
  • 1
    $\begingroup$ Right, they just forgot the $n$ in the statement. No worries! $\endgroup$
    – Dayton
    Sep 11, 2019 at 18:26

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.


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