I am trying to find following limit: $$\lim _{z\rightarrow \infty} \frac{z}{|z|}$$

where $z\in \mathbb{C}$. I have put the limit into wolfram alpha and the result is $1$. So I have been trying to prove it, but I seem to be confusing myself.

Note that $$\lim_{z\rightarrow \infty} \bigg |\frac{z}{|z|}\bigg| = 1 $$ and if $\lim _{z\rightarrow \infty} |f(z)| = c>0$ then also $\lim _{z\rightarrow \infty} f(z) = c$ since $|\cdot|$ is continous. So then the original limit should also be $1$.

However if I try using polar coordinates,

$$\lim_{z\rightarrow \infty} \frac{z}{|z|} = \lim_{r\rightarrow \infty} e^{i\theta}= e^{i\theta}$$ which depends on theta and so the limit shouldn't exist?

Q: What went wrong with the polar coordinate method?

  • $\begingroup$ Nothing went wrong with the polar method. This is in fact the right answer to the problem. $\endgroup$ – Tim B. Oct 27 '16 at 10:34
  • $\begingroup$ Then what went wrong with the other method? $\endgroup$ – fosho Oct 27 '16 at 10:35
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    $\begingroup$ |.| is continous so lim f(z) = c implies lim |f(z)] = |c| not the converse $\endgroup$ – InfiniteLooper Oct 27 '16 at 10:36
  • $\begingroup$ the one above the polar coordinate method $\endgroup$ – fosho Oct 27 '16 at 10:37
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    $\begingroup$ Your conclusion that $|f(z)|\to c>0$ implies $f(z)\to c$ doesn't follow from continuity. This is not at all what continuity says. It is still true however but only in the real numbers. $\endgroup$ – Tim B. Oct 27 '16 at 10:37

You have written

"if $\lim _{z\rightarrow \infty} |f(z)| = c>0$ then also $\lim _{z\rightarrow \infty} f(z) = c$ since $|\cdot|$ is continous."

But this is false ! An example is the limit

$$\lim _{z\rightarrow \infty} \frac{z}{|z|}$$.

This limit does not exist ! Nothing went wrong with the polar coordinate method !


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  • $\begingroup$ Alright, thanks a lot! $\endgroup$ – fosho Oct 27 '16 at 10:39
  • $\begingroup$ I think something important has been pointed out here. It looks for all the world at first glance that this limit does exist. So $+1$. $\endgroup$ – Teresa Lisbon Oct 27 '16 at 10:59

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