let $g(x)$ be a continuous function s.t. for each $x \ge 0$, $\lim_{x\to\infty} g(x)=L \ne 0$.

calculate the limit :

$$ \lim_{x\to\infty} \int_{0}^{1} g(xz) dz$$

SOLUTION ATTEMPT: I'm thinking about using the integration term by term Theorem, I know that because $g(x)$ is continuous in every closed interval $[0,b]$ it is integrable there. especially in $[0,1]$. so I can write:

$ \lim_{x\to\infty} \int_{0}^{1} g(xz) dz$=$\int_{0}^{1} \lim_{x\to\infty} g(xz) dz$

but didn't know how to continue from here. how do I integrate the right side? any kind of help would be appreciated.

  • $\begingroup$ I don't see any justification for you switching over the integral and the limit... $\endgroup$
    – 5xum
    Jan 15, 2016 at 11:31

1 Answer 1


Change variables $xz=t$ in the integral which becomes $$ \lim_{x\to\infty}\frac{1}{x}\int_0^x dt\ g(t) $$ and then by L'Hopital and using the fundamental theorem of calculus, the limit is equal to $L$.


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