# How to evaluate $\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{1-\sqrt{\sin 2x}}}{\pi-4x}$

I can not figure out why the limit does not exist ?

can anyone explain it ?

$$\lim_{x \to \frac{\pi}{4}} \frac{\sqrt{1-\sqrt{\sin 2x}}}{\pi-4x}$$ • There is something missing. What is the domain of study of your function? Jan 25 '16 at 8:55
• The limit doesn't exist because the value approached by the function differs depending on whether you get to $\pi/4$ from the left or from the right. In particular, note that the denominator is negative for $x>\pi/4$, making negative the whole function, whereas it's positive otherwise. Jan 25 '16 at 9:26
• Can you please tell me what book is that? Sep 4 '18 at 5:24

$$1-\sin2x=(\cos x-\sin x)^2$$
$$\sqrt{1-\sin2x}=|\sin x-\cos x|=\sqrt2\left|\sin\left(x-\dfrac\pi4\right)\right|$$
$=+(\sin x-\cos x)$ iff $x\ge\dfrac\pi4$
Considering $$A= \frac{\sqrt{1-\sqrt{\sin 2x}}}{\pi-4x}$$ start making life easier setting $x=\frac{\pi }{4}-y$. So $$A=\frac{\sqrt{1-\sqrt{\cos (2 y)}}}{4 y}$$ Now, use the fact that, for small $z$, $$\cos(z)=1-\frac{z^2}{2}+\frac{z^4}{24}+O\left(z^5\right)$$ which makes $$\cos(2y)=1-2 y^2+\frac{2 y^4}{3}+O\left(y^5\right)$$ Use the generalized binomial theorem to get $$\sqrt{\cos (2 y)}=1-y^2-\frac{y^4}{6}+O\left(y^5\right)$$ $$1-\sqrt{\cos (2 y)}=y^2+\frac{y^4}{6}+O\left(y^5\right)$$ $$\sqrt{1-\sqrt{\cos (2 y)}}=|y| \sqrt{1+\frac{y^2}6+\cdots}$$ So, by the end $$A=\frac 14 \frac{|y|} y \sqrt{1+\frac{y^2}6+\cdots}$$ So, depending if $y\to 0^+$, $A\to \frac 14$ and if $y\to 0^-$, $A\to -\frac 14$.