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I want to work out if the limit as $z$ approaches $0$ for $e^{1/z^4}$ exists and if not then why. I worked out that the left and right sided limits both equal to +∞ so I thought that was enough to conclude that the limit therefore existed.

But when I checked the solutions it said that the limit did not exist, because you get different limits when you approach ($0,0$) along different rays, ($x,y$) = ($at,bt$) as $t$ $\rightarrow$ $0$.

I don't understand this though can anybody help explain? What does it mean by the rays? Like gradients?

Edit $z$ is a complex number

sam
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  • Whati s $z$? How did a 2-d plane come into play with different variables $(x,y)$ ? – Rebellos Nov 11 '18 at 11:08
  • ah sorry $z$ is a complex number so i'd say $z$ = $x + iy$ – sam Nov 11 '18 at 11:22
  • a ray is just a straight line in the complex plane that contains the zero, that is, a ray is a path of the form ${rw:r\in\Bbb R}$ for some fixed $w\in\Bbb C\setminus{0}$ – Masacroso Nov 11 '18 at 11:25
  • so why does the limit change along different rays, wouldnt they all be zero as you approach zero? – sam Nov 11 '18 at 11:44

2 Answers2

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I don't think taking limits through rays is a good idea here. As $z$ approaches $0$ through the sequence $\{\frac 1 n\}$ the limit is $\infty$. If $z_n$ is a 4-th root of $\frac 1 {2n\pi i}$ the the limit through this sequence is $1$ (and $\{z_n\}$ does tend to $0$). Hence the limit does not exist.

Kavi Rama Murthy
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Hint:

Try checking limits for $z_n=\frac{1}{n}$ and for $z_n=\frac{1+i}{\sqrt2 \cdot n}$

Edit: You can also take a look at the plot

Jaroslaw Matlak
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