Given function $g$ continuous in line $[0,\infty)$...

Question

Given function $g$ continuous in line $[0,\infty)$, positive and strongly increasing that upholds $\lim_{t\to\infty} g(t) = L$. Calculate the limit: $$ \lim_{n\to\infty} \int\limits_1^2 g(nx) \, dx. $$

At the moment what I think that I should do is swap the $g(nx)$ with $L$ as it approaches infinity but from there on I am just not sure whats my next step is.

Answer 1

From the Lebesgue's monotone convergence theorem $$\lim\limits_{n\to \infty} \int\limits_1^2 g(nx) dx = \int\limits_1^2 \lim\limits_{n\to \infty} g(nx)dx = \int\limits_1^2 L dx = L . $$

Different approach. $$ g(n) =\int\limits_1^2g(n)dx\leq \int\limits_1^2 g(nx) dx \leq \int\limits_1^2 g(2n) dx = g(2n) $$ By the squezee theorem, we get the result.

Answer 2

This has nothing to do with monotonicity. If $g$ is continuous and $\lim_{t\to\infty}g(t)=L$ then $$g(nt)\to L\quad(n\to\infty)$$uniformly for $t\in[1,2]$, hence $\int_1^2 g(nt)\,dt\to L$.

Regarding a comment: Monotonicty is still irrelevant even if we assume $g$ is just measurable instead of continuous. If $g$ is measurable and $\lim_{t\to\infty}g(t)=L$ then the argument above applies without change to show $\int_1^2g(nt)\,dt\to L$.

Utterly and Completely Trivial Obvious Details: Suppose $\epsilon>0$. Choose $N$ so $|g(t)-L|<\epsilon$ for every $t>N$. Now if $n>N$ it follows that $|g(nt)-L|<\epsilon$ for every $t\in[1,2]$, since $nt>N$; this says precisely that $g(nt)\to L$ uniformly on $[1,2]$. Nothing to do with monotonicity or continuity.