Evaluating $\lim\limits_{x\to \frac {\pi}{2}} \tan (x) \log (\sin (x))$

Question

Evaluate $$\lim_{x\to \frac {\pi}{2}} \tan (x) \log (\sin (x)).$$

Any hints on this question? I didn't get any idea.

Answer 1

Try $$\lim_{x\to \frac {\pi}{2}}\tan x\cdot \log(\sin x)=\lim_{x\to \frac {\pi}{2}}\left(\frac{\log(\sin x)}{\dfrac{1}{\tan x}}\right)$$ And apply L'Hopital's to get$$\lim_{x\to \frac {\pi}{2}}\left(\dfrac{\cot x}{-\csc^2x}\right)=\lim_{x\to \frac {\pi}{2}}(-\cot x\cdot\sin^2x )$$

Answer 2

As Hanul Jeon commented, let $x=\frac \pi 2-y$, which makes $$\lim_{x\to \frac {\pi}{2}} \tan (x) \log (\sin (x))=\lim_{y\to 0}\frac{\log(\cos(y))}{\tan(y)} $$ Now, using Taylor series and composition of them $$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$ $$\log(\cos(y))=-\frac{y^2}{2}-\frac{y^4}{12}+O\left(y^6\right)$$ $$\tan(y)=y+\frac{y^3}{3}+\frac{2 y^5}{15}+O\left(y^6\right)$$ $$\frac{\log(\cos(y))}{\tan(y)}=\frac{-\frac{y^2}{2}-\frac{y^4}{12}+O\left(y^6\right) } {y+\frac{y^3}{3}+\frac{2 y^5}{15}+O\left(y^6\right) }=-\frac{y}{2}+\frac{y^3}{12}+O\left(y^5\right)$$

Just for the fun of it, consider $y=\frac \pi 6$ (corresponding to $x=\frac \pi 3$). The exact value of the expression is $$\sqrt{3} \log \left(\frac{\sqrt{3}}{2}\right)\approx -0.249140$$ while the approximation gives $$-\frac{\pi \left(216-\pi ^2\right)}{2592}\approx -0.249837$$