Using L'Hôpital's rule to find $\lim_{x\to\pi/2}(\tan x)(\ln \sin x)$ I have
$$\lim_{x\to\pi/2}(\tan x)(\ln \sin x)$$
And I need to solve it using L'Hôpital's rule. I can turn this limit around to get $\;\; (0/1)\cdot0$
But I don't see how to get $0/0$ to move on to the derivation.
 A: Rewrite:
$$\tan x\ln\sin x=\frac{\ln\sin x}{\cot x}\;.$$
This is a standard trick to deal with $0\cdot\infty$ forms: if $u\to 0$ and $v\to\infty$, then $\frac1v\to 0$, $uv=\frac{u}{1/v}$, and you can look at the $\frac00$ form instead. Sometimes it’s better to do the conversion the other way, to the $\frac{\infty}{\infty}$ form $\frac{v}{1/u}$.
A: Note also that you can rewrite $\;\color{blue}{\bf \tan x = \dfrac{\sin x}{\cos x}},\;$ 
 $$\lim_{x\to\pi/2}(\color{blue}{\bf \tan x})\cdot\ln (\sin x)\quad = \quad \lim_{x\to \pi/2} \frac{\color{blue}{\bf \sin x} \cdot \ln(\sin x)}{\color{blue}{\bf \cos x}},$$
which gives us the indeterminate form of $\;\large\frac 00,\,$ and ensures we can apply L'Hopital, as desired.
A: Another way of finding the limit is using the Taylor series
We have: $ u=x-\dfrac{\pi}{2} $ in order to obtain the limit as $u \to 0$.
As we used a change of variable, we must express everything in terms of the new variable.
So we have:
$$ L=\lim_{ x \to \frac{\pi}{2}}  \tan x \ln\left( \sin x  \right) =\lim_{ u \to  0}  \tan \left(\frac{\pi}{2}-u \right) \ln\left( \sin \left(\frac{\pi}{2}-u \right) \right)  $$
We recall some trigonometric identities:
$$ \sin\left(\frac{\pi}{2}-u \right)=\cos u  $$
$$ \cos\left(\frac{\pi}{2}-u \right)=\sin u  $$
$$ \tan\left(\frac{\pi}{2}-u \right)=\cot u  $$
Taking everything in terms of $ u $, we have
$$ L=\cot u \ln \left(\cos u\right)  $$
Using Taylor series:
$$\cos x= x-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!} - \dfrac{x^{6}}{6!}+...+\sum_{n=0}^{\infty}{ }  \frac{ \left( -1  \right)  ^{n}}{ \left(2n  \right)! } x^{2n} $$ 
$$\cos x= x- \dfrac{x^{2}}{2!}+O\left(x^{2} \right)  $$ 
$$\sin x= x- \dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!} - \dfrac{x^{7}}{7!}+...+\sum_{n=0}^{∞}{ }  \frac{ \left( -1  \right)  ^{n}}{ \left(2n+1 \right)! } x^{2n+1} $$ 
$$\sin x= x- \dfrac{x^{3}}{3!}+O\left(x^{3} \right)  $$ 
With this limit, we are left with:
$$ L\sim \lim_{ u \to 0} \dfrac{u- \dfrac{u^{2}}{2!}+O\left(u^{2} \right) }{u- \dfrac{u^{3}}{3!}+O\left(u^{3} \right)} \ln \left(u- \dfrac{u^{2}}{2!}+O\left(u^{2} \right) \right)=0 $$
A: I give here the limit by Taylor series just to show the simplicity of this method:
Let $h=x-\frac{\pi}{2}$ then $$\tan x \ln \sin x= \cot h \ln \cos h\sim_0\frac{1}{h}\ln(1-\frac{h^2}{2}+o(h^2))\sim_0-\frac{h}{2}$$ so the limit is $0$.
