$$\lim\limits_{n \to \infty} g(t)=L $$
$g$ is monotonic strictly increasing, continuous in $ [0,\infty)$ and positive function.
I need to find : $$\lim\limits_{n \to \infty} \int^2_1 g (nx) \,\mathrm{d}x $$
I have started with: $|g(t)-L|<\epsilon\>$
$L-\epsilon\>$$=<$$g(t)$$<=$$L+\epsilon\>$
$\int^2_1 (L-\epsilon\>) \,\mathrm{d}x$$=<$$\int^2_1 g (nx) \,\mathrm{d}x$$<=$$\int^2_1 (L+\epsilon\>) \,\mathrm{d}x$
I'm not sure if I'm in the right direction ...
You have using integral by substitution with $u = nx$: $$l_n = \int^2_1 g (nx) \,\mathrm{d}x = \frac{1}{n} \int_n^{2n} g (u) \,\mathrm{d}u $$
As $g$ is supposed to be strictly increasing, you have for $u \in [n, 2n]$ $g(n) \le g(u) \le g(2n)$. Therefore $$g(n) \le l_n \le g(2n)$$ by taking the integral of those inequalities over $[n,2n]$.
Knowing the hypothesis on $g$, $\lim\limits_{x \to \infty} g(x)$ exists (it can be $\infty$). Naming this limit $L$, you get $$\lim\limits_{n \to \infty} l_n = L$$