You can solve this problem just using definition of the limit.
Let us denote $I_n = \int\limits_0^2f(nx)\,dx = \frac1n\int\limits_0^{2n} f(x)dx$. Since $\lim\limits_{x\to\infty}f(x) = L<\infty$, for any $\varepsilon>0$ there is $x(\varepsilon)$ s.t. for all $x>x(\varepsilon)$: $ L-\varepsilon\leq f(x)\leq L+\varepsilon. $
Let us fix $\varepsilon>0$ and pick up any $n>x(\varepsilon)$, then $ I_n = \frac1n\int\limits_{0}^{x(\varepsilon)}f(x)dx+\frac1n\int\limits_{x(\varepsilon)}^{2n}f(x)dx\quad (1) $ and so $ \frac1nJ(\varepsilon)+\left(2-\frac{x(\varepsilon)}{n}\right)(L-\varepsilon)\leq I_n\leq \frac1nJ(\varepsilon)+\left(2-\frac{x(\varepsilon)}{n}\right)(L+\varepsilon)\quad (2) $ where $J(\varepsilon)= \int\limits_{0}^{x(\varepsilon)}f(x)dx$. In other words, $ \frac1n(J(\varepsilon)-(L+\varepsilon)x(\varepsilon))-2\varepsilon\leq I_n-2L\leq \frac1n(J(\varepsilon)-(L+\varepsilon)x(\varepsilon))+2\varepsilon. $
For any $\delta$ we pick up $\varepsilon<\delta$ and $N>\frac{3}{\delta}(J(\varepsilon)-(L+\varepsilon)x(\varepsilon))$, so for any $n\geq N$ we have $ |I_n-2L|\leq \delta, $ so $\lim\limits_{n\to\infty}I_n = 2L$.
Let me show how $(1)$ implies $(2)$: $ I_n = \frac1n J(\varepsilon)+\frac1n\int\limits_{x(\varepsilon)}^{2n}f(x)dx\leq \frac1n J(\varepsilon) +\frac{2n-x(\varepsilon)}{n}(L+\varepsilon)=\frac1nJ(\varepsilon)+\left(2-\frac{x(\varepsilon)}{n}\right)(L+\varepsilon). $ where the inequality holds because $f(x)\leq L+\varepsilon$ for all $x\geq x(\varepsilon)$. The other inequality in $(2)$ is obtained in a similar way since $f(x)\geq L-\varepsilon$ for all $x\geq x(\varepsilon)$.