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My teacher said without explaining that the limit of $\frac{e^{-\frac{1}{z^4}}}{z}$, $z\rightarrow 0$, where $z$ is a complex number doesn't exist. Why is this true?

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    Because it has an essential singularity at $z=0$. – 2017-01-08
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    Because the limit is different depending on how you approach $0$. – 2017-01-08
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    @Arthur, but why does the way you approach $z=0$ change the limit? – 2017-01-08
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    Try approaching along the positive real axis, and then try approaching from the direction of $1+i$. You will get very different results. – 2017-01-08
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    Not that I have a problem with you asking here, but did you ask your teacher, by any chance? – 2017-01-08

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Consider the sequences $$ a_n=\dfrac1n,\quad b_n=\dfrac1ne^{i\pi/4}. $$ They both converge to $0$, i.e. $$ \lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=0, $$ but $$ \lim_{n\to\infty}\dfrac{e^{-\dfrac{1}{a_n^4}}}{a_n}=\lim_{n\to\infty}ne^{-n^4}=0, $$ while $$ \lim_{n\to\infty}\dfrac{e^{-\dfrac{1}{b_n^4}}}{b_n}=e^{-i\pi/4}\lim_{n\to\infty}ne^{n^4}=\infty. $$

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The simplest way to prove this is using power series.

$$e^z = \sum_0^\infty \frac{z^n}{n!}$$

So

$$e^\frac{-1}{z^4} = \sum_0^\infty \frac{(-1)^n}{z^{4n}n!}$$

And

$$\frac{e^\frac{-1}{z^4}}{z} = \sum_0^\infty \frac{(-1)^n}{z^{4n+1}n!}$$

This certainly has no finite limit. It has an essential singularity at $0$ so by the Casorati-Weierstrass theorem it has no limit (not even $\infty$) at $0$.