1
$\begingroup$

Find $$\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$$


My attempt: ON THE basis of This post $$\lim_{x\to1}\tan\frac{\pi x}{4} =1,\quad \lim_{x\to1}\tan\frac{\pi x}{2}=\infty$$

$$\implies\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}= e^{\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}}$$

Now I need to solve $\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}$, but I don't know how to go on.

$\endgroup$
  • $\begingroup$ Could you explain why $$\lim_{x\to 1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}} = e^{\lim_{x\to 1}\left[\tan\frac{\pi x}{4}-1\right]\tan \frac{\pi x}{2}}$$? I don't understand that. $\endgroup$ – Matthew Leingang Oct 12 '17 at 12:28
  • $\begingroup$ @MatthewLeingang That is a standard result for functions like {f(x)}^g(x) $\endgroup$ – Kislay Tripathi Oct 12 '17 at 12:33
  • $\begingroup$ Maybe so. But where does it come from? $\endgroup$ – Matthew Leingang Oct 12 '17 at 12:35
  • 1
    $\begingroup$ I keep asking you to explain it, but you're just repeating it. Do you know why it's true? My point is that this might not be the most appropriate formula for the problem. $\endgroup$ – Matthew Leingang Oct 12 '17 at 12:48
  • 1
    $\begingroup$ Aside from the fact that you phrased your question in the imperative mood ("Please solve it."), which is considered rude in this community, your question is an XY problem. $\endgroup$ – Matthew Leingang Oct 12 '17 at 13:00
2
$\begingroup$

$$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}=\lim_{x\rightarrow1}\left(1+\tan\frac{\pi x}{4}-1\right)^{\frac{1}{\tan\frac{\pi x}{4}-1}\cdot\tan\frac{\pi x}{2}\left(\tan\frac{\pi x}{4}-1\right)}=$$ $$=e^{-\lim\limits_{x\rightarrow}\frac{2\tan\frac{\pi x}{4}}{1+\tan\frac{\pi x}{4}}}=e^{-1}=\frac{1}{e}.$$ I used $\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=-\frac{2\tan\alpha}{(\tan\alpha-1)(1+\tan\alpha)}.$

$\endgroup$
4
$\begingroup$

$$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}} = \lim_{x \to 1}e^ {{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}$$ Now we will check \begin{align*} \lim_{x \to 1} {{\tan\frac{\pi x}{2}}\ln\left(\tan\frac{\pi x}{4}\right)} &\stackrel{\text{(L'Hôpital's rule)}}{=} \lim_{x \to 1} \frac{-\frac{\pi}{4}\tan^2(\frac{x\pi}{2})\cos^2(\frac{x\pi}{2})}{\cos^2(\frac{x\pi}{4})\tan(\frac{x\pi}{4})} \\&= \frac{\frac{-1}{2}}{\frac{1}{2}} = -1 \ \end{align*}

and therefore: $\lim_{x \to 1} e^{{{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}} = e^{-1}$

$\endgroup$
  • 1
    $\begingroup$ That's the solution I was looking for. Good work. (PS: I reformatted, hope that's OK) $\endgroup$ – Matthew Leingang Oct 12 '17 at 12:46
2
$\begingroup$

Let $\tan\dfrac\pi4+1=u$ and using $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$

$$\lim_{x\to1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}}$$

$$=\lim_{u\to0}(1+u)^{\frac{2(u-1)}{1-(u-1)^2}}$$

$$=\left(\lim_{u\to0}(1+u)^{1/u}\right)^{\lim_{u\to0}\frac{2u(u-1)}{2u-u^2}}$$

Clearly, $\lim_{u\to0}(1+u)^{1/u}=e$ and $$\lim_{u\to0}\frac{2u(u-1)}{2u-u^2}=\lim_{u\to0}\frac{2(u-1)}{2-u}=?$$

$\endgroup$
2
$\begingroup$

Using your formula you have already covered the hard part. To proceed further just put $t=\tan (\pi x/4)$ so that $\tan(\pi x/2)=2t/(1-t^{2})$. The limit at the end of your question is $$\lim_{t\to 1}(t-1)\cdot\dfrac{2t}{1-t^{2}}=\lim_{t\to 1}-\frac{2t}{1+t}=-1$$ and the desired answer is $e^{-1}=1/e$.


By the way "not thinking about logics..." and "solving as many problems..." is not the way to beat competition because almost everyone is trying these easy approaches. You should go for "understanding basics" and this is the ability which separates the wheat from chaff in such competitions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.