Method for computing limit of a sin function as x tends to zero

Question

I have a question about computing $$ \lim_{x \to 0} \sin\left(\frac{\pi x}{4|x|}\right)$$ I found the limit of $\pi x$ and $4|x|$ seperately and ended with $\sin(\pi/4)$ which is equal to $1/\sqrt{2}$

Answer 1

We know that $\displaystyle\lim_{x \to 0} \sin \left ( \frac{\pi x}{4 |x|} \right )$ exists only if both the limit from the left and from the right exist as well.

$$\lim_{x \to 0^{-}} \sin \left ( \frac{\pi x}{4 |x|} \right ) = \lim_{x \to 0^{-}} \sin \left ( \frac{\pi x}{-4x} \right ) = -\sin \left ( \frac \pi 4 \right )$$

$$\lim_{x \to 0^{+}} \sin \left ( \frac{\pi x}{4 |x|} \right ) = \lim_{x \to 0^{+}} \sin \left ( \frac{\pi x}{4x} \right ) = \sin \left (\frac \pi 4 \right )$$

Hence the limit does not exist.

Regarding your cosine variation: $\displaystyle\lim_{x \to 0} \cos \left ( \frac{\pi x}{4 |x|} \right )$ exists because $\cos$ is an even function and therefore you have $\cos(-x) = \cos(x)$. In this case then the limit is $\displaystyle\cos \frac \pi 4 = \frac 1 {\sqrt{2}}$.