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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.

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Using the mean-value theorem, we have for some $\phi \in [\pi/4,x]$

$$\begin{align} \frac{x}{4x-\pi}\int_{\pi/4}^x \frac{\tan^2(\theta)}{\theta^2}\,d\theta&=\frac{x}{4x-\pi}\left(\frac{\tan^2(\phi)}{\phi^2}\right)(x-\pi/4)\\\\ &=\frac{x}{4}\left(\frac{\tan^2(\phi^2)}{\phi^2}\right)\\\\ &\to \frac{\pi}{16}\frac{1}{(\pi/4)^2}\,\,\text{as}\,\,x\to \pi/4\\\\ &=\frac{1}{\pi} \end{align}$$

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  • $\begingroup$ This is a stupid question: But what makes it legitimate to switch $$\int_{\pi/4}^x \frac{\tan^2(\theta)}{\theta^2}d\theta \quad \text{ for }\quad\left(\frac{\tan^2(\phi)}{\phi^2}\right)(x-\pi/4)$$ $\endgroup$ – Billy Rubina Dec 11 '16 at 6:11
  • $\begingroup$ That is: The first thing I'd think of would be evaluating the integral. $\endgroup$ – Billy Rubina Dec 11 '16 at 6:12
  • $\begingroup$ @OppaHilbertStyle, mean value theorem. $\endgroup$ – IAmNoOne Dec 11 '16 at 6:20
  • $\begingroup$ @Nameless I know that. But when I did calculus, the MVT seemed more a random curiosity than something that would make evaluation of integrals that easy. I'm quite surprised! $\endgroup$ – Billy Rubina Dec 11 '16 at 6:27
  • $\begingroup$ @Deepsea By "advanced" did you mean my accruing MSE reputation points? $\endgroup$ – Mark Viola Dec 11 '16 at 17:05
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In a very similar spirit with the above answers, we can look at part of quotient as a definition of derivative of the function $h(x) = \displaystyle \int_{0}^x \dfrac{\tan^2 \theta}{\theta^2}d\theta$. Thus the limit you are finding is $\displaystyle \lim_{x \to \frac{\pi}{4}} \dfrac{x}{4}\left(\dfrac{h(x)-h(\frac{\pi}{4})}{x-\frac{\pi}{4}}\right)$, and this limit is the derivative at $x = \dfrac{\pi}{4}$ of $h(x) $ times $\dfrac{\pi}{16}$,and using the F.T.C as pointed out by the second answer, you have $L = \dfrac{\pi}{16}h'(\frac{\pi}{4})= \dfrac{\pi}{16}\cdot \dfrac{16}{\pi^2} = \dfrac{1}{\pi}$.

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By the definition of derivatives and the fundamental theorem of calculus, $$ \lim_{x\to\pi/4}\frac{1}{x-\pi/4}\int_{\pi/4}^x\frac{\tan^2\theta}{\theta^2}\,dt=\frac{\tan^2(\pi/4)}{(\pi/4)^2}. $$ This means, by the product rule for limits, that your limit equals $$ \frac{\pi/4}{4}\cdot\frac{\tan^2(\pi/4)}{(\pi/4)^2}=\frac{1}{\pi}. $$

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