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For complex number $z$, how to prove the limit exits by definition

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

By def I've tried, but I got a difference always greater than 1..

Surb
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leave2014
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  • @Dietrich Burde: None of these related questions address the case where $z$ is taken to be a complex number. – J. J. Sep 11 '14 at 08:39
  • @J.J. OK, sorry. I thought the arguments would also work for complex numbers, and that this is a duplicate. – Dietrich Burde Sep 11 '14 at 08:45

1 Answers1

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L'Hopital's rule is a local statement; it concerns the behavior of functions near a particular point. The global issues (multivaluedness, branch cuts) are irrelevant.

$$\lim_{x \to 0} \frac{\sin(x)}{x} \stackrel{\text{de l'hopital}}{=} \lim_{x \to 0} \frac{cos(x)}{1}=1$$

UserX
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  • I just worry about, if the limit not exits, the L'hospital rule would not hold, thus, can we use it to prove the limit exist..? – leave2014 Sep 11 '14 at 09:00
  • I'm not sure, I'm inclined to No though. Sounds like circular logic to me. – UserX Sep 11 '14 at 09:13
  • ye... the derivative of $\sin x$ at $x=0$ require this limit exist, so, here we cannot use it to prove itself... – leave2014 Sep 11 '14 at 10:57