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Does

$\lim_{z\to 0} \frac{|z|}{z}$ exists?

If we look at real number x then $\lim_{x\to 0} \frac{|x|}{x}$ does not exists.

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    What do you think? It doesn't seem to exist, right? Can you verify this by taking two different paths to zero to get two different limits? (Like we do in the real case, by taking left and right limits) – 2017-01-10
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    No it wont exists...along real path it will be $\lim_{x-> 0} \frac{|x|}{x}$ whai has two different values – 2017-01-10
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    along Real path only limit becomes a problem – 2017-01-10
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    Perfect. So that's your answer, sewn up. – 2017-01-10
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    is |Z| differentiable at 0? – 2017-01-10
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    check if it satisfies the Cauchy Riemann equations at the point zero. $|z|= \sqrt {x^2+y^2}$, so you just have to check if a bunch of partial derivatives match at zero. – 2017-01-10
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    @MilanAmrutJoshi It isn't $|z|$ does not attain any non-real complex number (and is non-constant) hence cannot be complex differentiable anywhere. – 2017-01-10

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Bear in mind that $z=|z|e^{i\theta}$ where $\theta$ is the argument of $z$ so along each direction the limit will be $e^{-i\theta}$ and therefore the value is different along each half axis and the limit does not exist

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Presumably you want to know what happens to $|z|/z$ as $z\to 0$ thru non-real values. With $i^2=-1$ and any $n\in \mathbb N$ we have $|(i/n)|/(i/n)=-i$ and $|(-i/n)|/(-i/n)=+i,$ so no limit exists.

To put this another way: Let $f(z)=|z|/z$ when $z\ne 0.$ Let $z_n=i/n$ when $n$ is odd and $z_n=-i/n$ when $n$ is even. The sequence $(z_1,z_2,z_3,...)$ converges to $0$ but the sequence $(f(z_1),f(z_2),f(z_3),...)$ is $(-i,+i,-i,+i,...)$ which does not converge.