$\lim_{x \to \frac{\pi}{2}} (\pi - 2x) \tan x = $

Question

$\lim_{x \to \frac{\pi}{2}} (\pi - 2x) \tan x = $ can this be solved by something like this :

$\pi - 2x = t$

$x = \frac{\pi - t}{2}$

$x \to \frac{\pi}{2}$

$\frac{\pi - t}{2} \to \frac{\pi}{2}$

$t \to 0$

without derivative nor $\cot (\frac{\pi}{2} - x)$

Answer 1

Yes, with your substitution $\pi-2x=t$, we have $$\tan x=\tan\frac{\pi-t}{2}=\tan\left(\frac \pi 2-\frac t2\right)=\cot\frac t2 $$ the limit becomes $$\lim_{t\to 0}t\cot{\frac t2}=\lim_{t\to 0}t\frac{\cos\frac t2}{\sin \frac t2}=2 $$ using $\lim_{y\to 0}\frac{y}{\sin y}=1$ for $y=\frac t2$.

Answer 2

Starting from @bjorn93's answer, if you want more than the limit, use the Taylor expansion of $\cot(x)$ to get $$t \cot \left(\frac{t}{2}\right)=2-\frac{t^2}{6}+O\left(t^4\right)$$ which, for sure, shows the limit but also how it is approached when $t\to 0$.

It also gives you a good approximation : trying for $t=\frac \pi 6$ (this is far away from $0$), the exact value is $\frac{2+\sqrt{3}}{6} \approx 1.95410$ while the truncated expansion gives $2-\frac{\pi ^2}{216}\approx 1.95431$.