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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.

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Why doesn't it exist here? Is it because it has more than 1 limit point? – David Thompson Nov 1 '12 at 20:39
@DavidThompson Exactly. Limit point depends on the path you take, which implies the limit does not exist. The standards for complex limit's existence are much more demanding than for the real one, and this problem is an excellent example why. – gt6989b Nov 1 '12 at 20:43
Thank you. So if I say "z=z* in the reals, the limit of z*/z as z goes to 0 would be 1. But from the imaginary axis z=-z*, the limit of z*/z would be -1. Since there are two limit points, you know it cannot have an actual limit and hence it does not exist." Would that be a sufficient proof? No epsilon/delta argument? – David Thompson Nov 1 '12 at 20:54
@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

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