In this answer it is proved that if $f, g$ are Riemann integrable on $[0,T]$ and $g$ is periodic with period $T$ then $$\lim_{n\to\infty} \int_{0}^{T}f(x)g(nx)\,dx=\frac{1}{T}\left(\int_{0}^{T}f(x)\,dx\right)\left(\int_{0}^{T}g(x)\,dx\right)$$ For your problem we have $$f(x) =\sin x, g(x) =\frac{1}{1+\cos^2x}$$ and hence the desired limit is $$\frac{1}{\pi}\int_{0}^{\pi}\sin x\, dx\int_{0}^{\pi}\frac{dx}{1+\cos^2x}=\frac{4}{\pi}\int_{0}^{\pi}\frac{dx}{3+\cos x}=\frac{4}{\pi}\cdot\frac{\pi}{2\sqrt{2}}=\sqrt{2}$$
It can be solved using differentiation under the integral sign. Consider the following integral:
\begin{equation}
I(t)=\int\limits_{-\infty}^{+\infty} \frac{\cos(tx)}{x^{2}+k} \,dx = 2\int\limits_{0}^{+\infty} \frac{\cos(tx)}{x^{2}+k} \,dx
\end{equation}
for any positive real $t$ and $k$. The first derivative with respect to $t$ is:
\begin{equation}
I'(t)= -2\int\limits_{0}^{+\infty} \frac{x\sin(tx)}{x^{2}+k} \,dx
\end{equation}
\begin{equation}
\Leftrightarrow \hspace{.3cm}I'(t)= -2\int\limits_{0}^{+\infty} \frac{x^{2}\sin(tx)}{x(x^{2}+k)} \,dx
\end{equation}
\begin{equation}
\Leftrightarrow \hspace{.3cm}I'(t)= -2\int\limits_{0}^{+\infty} \frac{(x^{2}+k-k)\sin(tx)}{x(x^{2}+k)} \,dx
\end{equation}
\begin{equation}
\Leftrightarrow \hspace{.3cm}I'(t)= -2\int\limits_{0}^{+\infty} \frac{\sin(tx)}{x} \,dx +2k\int\limits_{0}^{+\infty} \frac{\sin(tx)}{x(x^{2}+k)} \,dx
\end{equation}
The first one is just the sine integral as $x\rightarrow \infty$ and it is known to converge to $\frac{\pi}{2}$. Thus:
\begin{equation}
I'(t)= 2k\int\limits_{0}^{+\infty} \frac{\sin(tx)}{x(x^{2}+k)} \,dx -\pi
\end{equation}
Differentiating once more with respect to $t$ yields:
\begin{equation}
I''(t)= 2k\int\limits_{0}^{+\infty} \frac{\cos(tx)}{x^{2}+k} \,dx
\end{equation}
\begin{equation}
\Leftrightarrow \hspace{.3cm}I''(t)-kI(t)=0
\end{equation}
The general solution to the ODE is:
\begin{equation}
I(t)=c_{1}e^{\sqrt{k}t}+c_{2}e^{-\sqrt{k}t}
\end{equation}
Plugging some conditions $\left(I(t=0) \,\,\text{and}\,\, I'(t=0)\right)$ allows you to find that $c_{1}=0$ and that $c_{2}=\frac{\pi}{\sqrt{k}}$. Then:
\begin{equation}
\boxed{\int\limits_{-\infty}^{+\infty} \frac{\cos(tx)}{x^{2}+k} \,dx = \frac{\pi}{\sqrt{k}}e^{-\sqrt{k}t}}
\end{equation}
for positive real values of $t$ and $k$. If you plug $t=2$ and $k=4$, you obtain the desired result.
Best Answer
When applying l'hospital, why didn't you derive $\left(x-1\right)^3$, that'd make de $1/3$ appear ?
It'd give
$$ \lim\limits_{x \rightarrow 1}\frac{\ln^2\left(x\right)}{3\left(x-1\right)^2} $$