Right and left limits of $\frac{\sqrt{\sin x^2}}x$ at 0

Question

Can someone help me to find left and right limits of $\frac{\sqrt{\sin x^2}}x$ at 0? When I draw the graph it is clear that left and right limits are 1 and $-1$, but don't know to show that in proper way.

Answer 1

Note that since $\sqrt {x^2}=|x|$, we have that

• for $x>0$

$$\frac{\sqrt{\sin x^2}}x= \sqrt{\frac{\sin x^2}{x^2}}\to\sqrt 1=1$$

• for $x<0$

$$\frac{\sqrt{\sin x^2}}x= -\sqrt{\frac{\sin x^2}{x^2}}\to-\sqrt 1=-1$$

Answer 2

$\lim_{x\to 0}\frac{\sqrt{\sin x^2}}x \simeq \lim_{x\to 0}\frac{\sqrt{x^2}}x=\lim_{x\to 0}\frac{|x|}x$

For $x=0^+ : \lim_{x\to 0}\frac{|x|}x=\lim_{x\to 0}\frac{x}x=1$

For $x=0^- : \lim_{x\to 0}\frac{|x|}x=\lim_{x\to 0}\frac{-x}x=-1$