Limits with Taylor series

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

Hence, $$\lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos^2x}{\ln\sin{x}}=\lim_{t\rightarrow0}\frac{\sin^2t}{\ln\cos{t}}=\lim_{t\rightarrow0}\left(\frac{-2\sin^2\frac{t}{2}}{\ln\left(1-2\sin^2\frac{t}{2}\right)}\cdot\frac{\sin^2t}{-2\sin^2\frac{t}{2}}\right)=-2$$ because $\sin{t}=t-\frac{t^3}{6}+...$

For the first

consider:

$$\lim_{x\to \pi/2}\frac{\cos^2 x}{\log \sin x }=\lim_{y\to 0}\frac{\sin^2 y}{\log \cos y }$$

since:

$\sin^2 y=y^2+o(y^2)$

$\cos y=1-\frac{y^2}{2}+o(y^2)$

$\log \cos y=\log\left(1-\frac{y^2}{2}+o(y^2)\right)=-\frac{y^2}{2}+o(y^2)$

we have

$$\frac{\sin^2 y}{\log \cos y }=\frac{y^2+o(y^2)}{-\frac{y^2}{2}+o(y^2)}=\frac{1+o(1)}{-\frac{1}{2}+o(1)}\to -2$$

For the second

$$\arctan x=\frac{\pi}{2}-\arctan\frac1x$$

$$x\frac{\pi}{2}- x\arctan(x)=x\arctan\frac1x=x\cdot\frac1x+o(1)\to 1$$

NOTE

For both limits is not strictly necessary use Taylor's expansion, It suffices use standard limits.