$g$ is monotonic strictly increasing, continuous in $ [0,\infty)$ and positive function. $$\lim\limits_{n \to \infty} g(t)=L $$
 $g$ 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: 
 $|g(t)-L|&lt;\epsilon\&gt;$

$L-\epsilon\&gt;$$=&lt;$$g(t)$$&lt;=$$L+\epsilon\&gt;$

$\int^2_1 (L-\epsilon\&gt;) \,\mathrm{d}x$$=&lt;$$\int^2_1 g (nx) \,\mathrm{d}x$$&lt;=$$\int^2_1 (L+\epsilon\&gt;) \,\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
$$\lim\limits_{n \to \infty} l_n = L$$
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
$$\lim\limits_{n \to \infty} l_n = L$$