The limit of $\frac{\sqrt{1+\cos 2x}}{\sqrt{\pi}-\sqrt{2x}}$ as $x\to \pi/2+0$

Question

I cannot find the one-sided limit $$\lim_{x\rightarrow \pi/2+0} \frac{\sqrt{1+\cos 2x}}{\sqrt{\pi}-\sqrt{2x}}$$ How to get rid of the square root here? (without L'Hopital's rule)

Answer 1

We have $$\eqalign{\frac{\sqrt{1+\cos2x}}{\sqrt\pi-\sqrt{2x}} &=-\sqrt2\cos x\,\frac{\sqrt\pi+\sqrt{2x}}{\pi-2x}\cr &=-\sqrt2(\sqrt\pi+\sqrt{2x})\frac{\sin(\frac\pi2-x)}{2(\frac\pi2-x)}\cr &\to-\sqrt{2\pi}\ .\cr}$$

Answer 2

You could also use Taylor expansions around $x=\frac \pi 2$ $$\cos(2x)=-1+2 \left(x-\frac{\pi }{2}\right)^2+O\left(\left(x-\frac{\pi }{2}\right)^4\right)$$ So $$1+\cos(2x)=2 \left(x-\frac{\pi }{2}\right)^2+\cdots$$ $$\sqrt{1+\cos(2x)}=\sqrt2 \left(x-\frac{\pi }{2}\right)+\cdots$$

I am sure that you can take from here.

Answer 3

Set, for simplicity, $2x=\pi+t$. Then your limit is \begin{align} \lim_{t\to0^+}\frac{\sqrt{1+\cos(\pi+t)}}{\sqrt{\pi}-\sqrt{\pi+t}} &= \lim_{t\to0^+}\frac{\sqrt{1+\cos(\pi+t)}}{\pi-\pi-t} (\sqrt{\pi}+\sqrt{\pi+t})\\[6px] &= \lim_{t\to0^+}-(\sqrt{\pi}+\sqrt{\pi+t})\sqrt{\frac{1-\cos t}{t^2}} \end{align} The function under the big square root sign has a known limit.