$\lim_{x \to \pi/4} \frac{x}{4x-\pi} \int_{\pi/4}^x \frac{tan^2\theta}{\theta^2} d\theta $ I need help solving this limit:
$\lim_{x \to \pi/4} \frac{x}{4x-\pi} \int_{\pi/4}^x \frac{tan^2\theta}{\theta^2} d\theta $
The limit is solvable using l'hopital's rule to get $\frac{64}{\pi^2}$, but I need to see if it is possible to do so without using it. This problem was presented to me by my friend who is taking Calculus 1 so nothing beyond simple integrals and derivatives if possible.

Using the mean-value theorem, we have for some $\phi \in [\pi/4,x]$
$$\begin{align}
\frac{x}{4x-\pi}\int_{\pi/4}^x \frac{\tan^2(\theta)}{\theta^2}\,d\theta&=\frac{x}{4x-\pi}\left(\frac{\tan^2(\phi)}{\phi^2}\right)(x-\pi/4)\\\\
&=\frac{x}{4}\left(\frac{\tan^2(\phi^2)}{\phi^2}\right)\\\\
&\to \frac{\pi}{16}\frac{1}{(\pi/4)^2}\,\,\text{as}\,\,x\to \pi/4\\\\
&=\frac{1}{\pi}
\end{align}$$
