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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.

Michael Rozenberg
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Kislay Tripathi
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  • 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. – Matthew Leingang Oct 12 '17 at 12:28
  • @MatthewLeingang That is a standard result for functions like {f(x)}^g(x) – Kislay Tripathi Oct 12 '17 at 12:33
  • Maybe so. But where does it come from? – Matthew Leingang Oct 12 '17 at 12:35
  • Go ahead and precede commons three-letter functions with a \ so they appear upright, like \lim $\lim$ and \tan $\tan$. – gen-ℤ ready to perish Oct 12 '17 at 12:38
  • @MatthewLeingang The standard result is for lim${x\longrightarrow a}$$\left{ f(x)\right} $$^{g(x)}$if lim${x\longrightarrow a}${f(x)}=1 and lim${x\longrightarrow a}$g(x)= $\infty$$\Longrightarrow$ lim${x\longrightarrow a}$$\left{ f(x)\right} $$^{g(x)}$ =e$^{lim_{x\longrightarrow a}{f(x)-1}g(x)}$ . please do not downvote my question. tell me what is wrong first. – Kislay Tripathi Oct 12 '17 at 12:45
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    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. – Matthew Leingang Oct 12 '17 at 12:48
  • @MatthewLeingang ok , so the truth is I do not know. I am preparing for a competition . i don't have enough time to think about the logics behind the formulas. i am only worried about solving as many question as possible.i hope you understand it.Proof of the result is not given in my book – Kislay Tripathi Oct 12 '17 at 12:52
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    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. – Matthew Leingang Oct 12 '17 at 13:00
  • @MatthewLeingang In my community it is said that Rod is god .You have the rod .No more explanations – Kislay Tripathi Oct 12 '17 at 13:05
  • @MatthewLeingang This is not XY problem.I asked help for solving $\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}$ Ok now see

    $\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\tan\frac{\pi x}{2}$= $\lim_{x\to1}\left[\tan\frac{\pi x}{4}-1\right]\frac{2tan\frac{\Pi x}{4}}{(1+tan\frac{\Pi x}{4})(1-tan\frac{\Pi x}{4})}=lim_{x\longrightarrow1}\frac{-2tan\frac{\Pi x}{4}}{1+tan\frac{\Pi x}{4}}$$\Longrightarrow$$\frac{-2}{2}$= -1$\Longrightarrow$Answer is $\frac{1}{e}$. So i was right in my question. I used, please because I need the answer

     – Kislay Tripathi Oct 12 '17 at 13:43
  • Since i have cleared everything will everyone be enough kind to take their downvotes back – Kislay Tripathi Oct 12 '17 at 13:56
  • Re: Proof of the result is not given in my book. Probably you could improve the question by explicitly saying the source instead of just saying my book. And you could also clearly state the result you are using in your attempt also in the post, not just in comments. – Martin Sleziak Dec 01 '17 at 05:17
  • Perhaps you could also quote this post in connection with that result: Regarding $e$ in $\lim\limits_{x \to a}{[\phi(x)]^{\psi(x)}} = e^{\lim\limits_{x \to a}{[\phi(x) - 1]\psi(x)}}$. – Martin Sleziak Dec 01 '17 at 05:20

4 Answers4

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$$\lim_{x\rightarrow1}\left(\tan\frac{\pi x}{4}\right)^{\tan\frac{\pi x}{2}} = \lim_{x \to 1}e^ {{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}$$ 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: $\lim_{x \to 1} e^{{{\tan\frac{\pi x}{2}}\ln(\tan\frac{\pi x}{4})}} = e^{-1}$

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

Michael Rozenberg
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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 $$\lim_{u\to0}\frac{2u(u-1)}{2u-u^2}=\lim_{u\to0}\frac{2(u-1)}{2-u}=?$$

lab bhattacharjee
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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.

Paramanand Singh
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