# How do we evaluate the limits: $\lim_{x \to 0^{+,-} } \frac{\sin\ 2x}{|\sin\ 2x|}, \lim_{x\to \frac{π}{2}^{+,-} } \frac{\sin\ 2x}{|\sin\ 2x|}$?

How do we evaluate the limits: $$\lim_{x \to 0^{+,-} } \frac{\sin\ 2x}{|\sin\ 2x|}, \lim_{x\to \frac{π}{2}^{+,-} } \frac{\sin\ 2x}{|\sin\ 2x|}$$

if we have absolute value at the denominator?

An idea:

$$\frac{\sin 2x}{|\sin 2x|}=\begin{cases}\frac{\sin 2x}{\sin2x}\;,\;\;x>0\;\;\text{and close to zero, say}\;\;|x|<10^{-3}\\{}\\\frac{\sin2x}{-\sin2x}\;,\;\;x<0\;\;\text{and close to zero, say}\;\;|x|<10^{-3}\end{cases}$$

and now evaluate both one-sided limits.

For the second one things are pretty similar as above. Try it.

• This does not hold for $x = \frac{3}{4}\pi$ for example. – GenericNickname Nov 23 '14 at 13:35
• Indeed so, @GenericNickname. Thanks, answer edited now. – Timbuc Nov 23 '14 at 13:37

Hint:

When $x< 0$ meaning limit from $-$, left side.

What is $|x| = ?$ equal to? Keep in mind $|x| > 0$ for all real $x$.

For $|\sin(2x)|$ when $\sin(2x) < 0$ we get $|\sin(2x)| = -\sin(2x)$

For values close to $x = 0$, approximately $-1 < x < 0$ 4th quadrant, $\sin(2x) < 0$

Try it.