# Find the limit of $\frac{\bar{z}}{z}$ as $z$ goes to $0$.

I put it in exponential form to get $\dfrac{re^{-i \theta}}{re^{i \theta}}$ but I think I'll get $\frac{0}{0}$ which isn't defined and isn't a good enough proof to say it doesn't have a limit.

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You have to decide how $z$ goes to $0$. In terms of $r$ and $x$ for example. Then you can take the limit as usual. –  mick Nov 1 '12 at 20:25

$$z=re^{it}\,\,,\,r>0\,,\,r\in\Bbb R\,\,,\,t\in [0,2\pi]\Longrightarrow\frac{\overline z}{z}=e^{-2it}$$

From the above it follows that the limit depends on the angle $\,t\,$ and, thus, it doesn't exist.

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Can you explain further please? –  David Thompson Nov 1 '12 at 20:40
I'm not sure...if $\,z=re^{it}\,$ , then $\,z\to 0\Longleftrightarrow r\to 0\,$, thus in the above we see that taking the limit we get the constant $\,e^{-2it}\,$ , which is always different for different values of $\,t\in [0,2\pi)\,$... –  DonAntonio Nov 1 '12 at 20:49
@DavidThompson For a limit to exist, it has to have the same value, no matter which direction you approach the limit from. This shows that the function approaches a different value depending on $t$. –  Navin Nov 2 '12 at 1:45
Consider approaching $0$ from the real axis, then $z \in \mathbb{R}$, so $z = \bar{z}$, hence the limit is $1$.
Now approach from the imaginary axis, now $z = -\bar{z}$ so limit is $-1$. Thus, it does not exist.
@DavidThompson Yes, sufficient to show there are subsequences converging to multiple limit points. It's akin to the real sequence $0,1,0,1,0,1,...$ which also diverges for the same reason - 2 subsequences converging to distinct limit points... –  gt6989b Nov 1 '12 at 21:00