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Can someone help me to calculate the following limit?

$$\lim_{x\to a} \Big(2-\frac{x}{a}\Big)^{\tan\dfrac{\pi x}{2a}}$$

Thank you.

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    No idea what you are asking. – 2017-01-27
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    find Limit x tends a (2-(x/a))^tan(pi*x/2*a) does this help? – 2017-01-27
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    Yeah, still no. The question is completely unreadable. – 2017-01-27
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    limx→a (2−(x/a))^tan(πx/2a) – 2017-01-27
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    Do you mean $\lim_{x\rightarrow 0} (2-\frac{x}{a})^{\tan (\frac{\pi x}{2a})}$? – 2017-01-27
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    yeah but x tend to a – 2017-01-27
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    I think because of new she has some mistakes. If someone can help. Please help. – 2017-01-27
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    @KanwaljitSingh I help people that show even a modest attempt at being understandable. This user clearly posted a question which was *obviously* unreadable, so I see no reason why I should put in all the work. – 2017-01-27
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    @5xum I think now question is clear. So if you can please help. – 2017-01-27
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    Nitisha: I wrote the limit as \lim_{x\rightarrow a} (2-\frac{x}{a})^{\tan (\frac{\pi x}{2a})}. In order to get the nice formulas, you have to put such expressions between dollar signs. This code is called mathjax, it's good to get familiar with this if you aim to ask more questions or give answers. – 2017-01-27
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    thank you one and all, I'll make sure not to repeat this mistake again. – 2017-01-27
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    I believe my answer deserves an upvote after the editing, advice and general hardship. Be sure to accept an answer to mark the question as answered. – 2017-01-27
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    I upvoted but it's not showing up publicly due to lack of required reputation. thanks so much – 2017-01-27
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    .....Brilliant? – 2017-01-28

2 Answers 2

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Consider $\lim_{x\rightarrow a} e^{\ln((2-\frac{x}{a})^{\tan (\frac{\pi x}{2a})})}=\lim_{x\rightarrow a} e^{\tan(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}$.

Clearly it suffices to calculate the limit $$\lim_{x\rightarrow a} {\tan(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}.$$ We can do this by using L'Hopitals rule.

\begin{eqnarray} \lim_{x\rightarrow a} {\tan(\frac{\pi x}{2a})\ln(2-\frac{x}{a})} &=& \lim_{x\rightarrow a} \frac{\sin(\frac{\pi x}{2a})\ln(2-\frac{x}{a})}{\cos(\frac{\pi x}{2a})}\\ & \overset{\mathrm{H}}{=}& \lim_{x\rightarrow a} \frac{\frac{\pi}{2a}\cos(\frac{\pi x}{2a})\ln(2-\frac{x}{a})-\frac{1}{a}\frac{1}{2-\frac{x}{a}}\sin(\frac{\pi x}{2a})}{-\frac{\pi}{2a}\sin(\frac{\pi x}{2a})}\\ &=& \frac{2}{\pi}. \end{eqnarray}

Hence the desired limit is $e^{\frac{2}{\pi}}$.

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Say $2-\dfrac{x}{a}=1+\dfrac{1}{y}$ then $y=\dfrac{a}{a-x}$ and $y\to\infty$ as $x\to a$. With this substitution we have $$\tan(\frac{\pi x}{2a})=\cot(\frac{\pi}{2}-\frac{\pi x}{2a})=\cot(\frac{\pi}{2}-\frac{\pi x}{2a})=\cot\frac{\pi}{2y}$$ so $$\lim_{x\to a} (2-\frac{x}{a})^{\tan (\frac{\pi x}{2a})}=\lim_{y\to\infty}\Big[(1+\dfrac{1}{y})^y\Big]^{\frac1y\cot\frac{\pi}{2y}}$$ But $$\lim_{y\to\infty}\frac1y\cot\frac{\pi}{2y}=\lim_{t\to0}\frac{t}{\tan\frac{\pi t}{2}}=\frac{2}{\pi}$$ Thus thee limit is $\color{red}{e^\frac{2}{\pi}}$.

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    Clever substitution, elegantly done. – 2017-01-27
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    @Mathematician42 Thanks. – 2017-01-27
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    Actually, the more I see the elegant introduction of $e$ into the limit, the more I start liking your answer. Creative solution, I doubt the unexperienced like such solutions as that cannot be thaught. – 2017-01-27
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    @Mathematician42 Exercise makes experience. – 2017-01-27