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$\lim_{z \to 0} \frac{ \sin |z| }{z} $. I claim this limit does not exist.

My try: I think, but I am not sure if I am allowed to do this here, we can find the limits from the left and from the right:

$$ \lim_{z \to 0^+} \frac{ \sin z }{z} = 1 $$

$$ \lim_{z \to 0^-} \frac{ \sin (-z) }{z} = -1 $$

So, the limit does not exist. Is there another way to show this limit does not exist?

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  • $\begingroup$ Right and left limits are a thing on the real line. But you are working in the complex plane. $\endgroup$ – Henno Brandsma Sep 26 '15 at 5:30
  • $\begingroup$ @HennoBrandsma You can take the limit as $z$ goes to $0$ along the real line in the complex plane ... $\endgroup$ – Zubin Mukerjee Sep 26 '15 at 5:43
  • $\begingroup$ Then he should say that (saying that he restricts to the real axis). Then prove a statement that if both of those are different, the limit does not exist. This should still be proved. $\endgroup$ – Henno Brandsma Sep 26 '15 at 5:45
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As mentioned in the comments, you "wrongly" showed the non-existence of the real limit (i.e., when the real number $z$ tends to $0$), whereas the question is about the complex limit. Nevertheless, this implies the non-existence of the complex limit as well: If the complex limit existed ($L$, say), then for all $\epsilon>0$ there'd be $\delta>0$ such $0<|z|<\delta$ implies $|L-\frac{\sin|z|}{z}|<\epsilon$. Already your consideration of positive and negative real numbers shows that we can always find $z\approx 0$ with $f(x)\approx 1$ and others with $f(z)\approx -1$, so for any $\epsilon<2$ no such $\delta$ can exist.

You could also try to find different limits along other paths to $0$, or exhibit a single complex sequence where the limit does not exist.

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