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Evaluate the limit without L'Hospital's Rule and Taylor Series: $${\displaystyle \lim_{x\to1}\left(x-1\right)\tan\left(\frac{\pi x}2\right)}$$

I can't seem to go find a substitution that works for this limit. I tried with $u=x-1$, but do not know where to go from there.

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    $\begingroup$ Try to use $y=1-x$ then make the $tan$ becomes $\frac{sin}{cos}$. I'm too lazy to make it as a long solution but if you know $$\lim_{x\rightarrow 0} \frac{sin x}{x}=1$$ then you can finish it. $\endgroup$ – ElfHog Feb 2 '18 at 3:35
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$$ \begin{aligned} &{\displaystyle \lim_{x\to1}\left(x-1\right)\tan\left(\frac{\pi x}2\right)}\to\small{\begin{bmatrix} &t=x-1&\\ &t\to0& \end{bmatrix}}\to{\displaystyle \lim_{t\to0}t\cdot\tan\left(\frac{\pi t}2+\frac{\pi}2\right)}=-{\displaystyle\lim_{t\to0}t\cdot\cot\left(\frac{\pi t}2\right)}=\\ \\ &=-{\displaystyle\lim_{t\to0}t\frac{\cos\left(\frac{\pi t}2\right)}{\sin\left(\frac{\pi t}2\right)}}=-{\displaystyle\lim_{t\to0}\frac{\cos\left(\frac{\pi t}2\right)}{\sin\left(\frac{\pi t}2\right)}\cdot\frac{\frac{\pi}{2}t}{\frac{\pi}2}}=-{\displaystyle\lim_{t\to0}\frac{2\cos\left(\frac{\pi t}2\right)}{\pi}}\cdot1=-\frac{2}{\pi}\,. \end{aligned}$$

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    $\begingroup$ +1) Nice. For the OP: Is it clear how the answer went from that tangent term to the cotangent term in the next = sign? Use sum formula for tangent and multiply through by $cos\frac{\pi}{2}$ $\endgroup$ – imranfat Feb 2 '18 at 3:40
  • $\begingroup$ You could just use $\lim_{y \to 0} \frac{\tan y}{y} = \lim_{y \to 0} \frac{y}{\tan y} =1$ directly, of course. $\endgroup$ – Deepak Feb 2 '18 at 8:12
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    $\begingroup$ @Deepak I wrote the solution in this way to make it clearer, so that every step is accounted for. $\endgroup$ – Denis28 Feb 2 '18 at 8:31

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