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.