1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
7
$\begingroup$

$$ \begin{aligned} &{\displaystyle \lim_{x\to1}\left(x-1\right)\tan\left(\frac{\pi x}2\right)}\to\small{\begin{bmatrix} &t=x-1&\\ &t\to0& \end{bmatrix}}\to{\displaystyle \lim_{t\to0}t\cdot\tan\left(\frac{\pi t}2+\frac{\pi}2\right)}=-{\displaystyle\lim_{t\to0}t\cdot\cot\left(\frac{\pi t}2\right)}=\\ \\ &=-{\displaystyle\lim_{t\to0}t\frac{\cos\left(\frac{\pi t}2\right)}{\sin\left(\frac{\pi t}2\right)}}=-{\displaystyle\lim_{t\to0}\frac{\cos\left(\frac{\pi t}2\right)}{\sin\left(\frac{\pi t}2\right)}\cdot\frac{\frac{\pi}{2}t}{\frac{\pi}2}}=-{\displaystyle\lim_{t\to0}\frac{2\cos\left(\frac{\pi t}2\right)}{\pi}}\cdot1=-\frac{2}{\pi}\,. \end{aligned}$$

$\endgroup$
  • 2
    $\begingroup$ +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
  • 4
    $\begingroup$ @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

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.