Limits with Taylor series I'm stuck computing these two limits using Taylor series. 
The first is
1) $$\lim_{x\to \pi/2}\frac{\cos^2(x)}{\log[\sin(x)]}$$
and the second one is 
2) $$\lim_{x\to \infty}\frac{x(\pi/2) - x\arctan(x)}{1}$$
I tried using the already known Taylor series, in both the two limits (and I also tried using higher orders) but I don't seem to get anywhere.
For example,  in the second limit doing a sobstitution that seems obvious to me 
 $$\{x\to \frac{1}{t}$$
 leads to 
$$\lim_{t\to 0} {\frac{1}{t}\left [\frac{\pi}{2} -\frac {1}{t} +\frac{3}{t^3} -\frac{5}{t^5} +\frac{o\left(t^5\right)}{1}\right]}$$
that is the same form I had at the beginning. Using an higher grade of the Taylor expansion doesn't change this form. 
I'm probably missing something. Can someone please explain to me what has to be done or what I'm doing wrong? 
 A: 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 $\sin{t}=t-\frac{t^3}{6}+...$
A: 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.
A: Alternatively, using Lopital's rule:
$$L=\lim_{x\to\frac{\pi}{2}} \frac{-2\sin x \cos x}{\frac{\cos x}{\sin x}}=-2\lim_{x\to\frac{\pi}{2}} \sin^2 x=-2.$$
$$L=\lim_{x\to\infty} \frac{\frac{\pi}{2}-\arctan x}{\frac{1}{x}}=\lim_{x\to\infty} \frac{-\frac{1}{1+x^2}}{-\frac{1}{x^2}}=1.$$
