# Show limit does not exist $\lim_{z \to 0}e^{\frac{1}{z}}$

How would you formally show that $\lim_{z \to 0}e^{\frac{1}{z}}$ does not exist?

I think you just show that the limit along two different paths towards the origin are different such as along $z=x$ and $z=iy$ but I don't know how to show it.

• that is the idea. How, about setting the "imaginary" portion to 0, and compare the left hand limit to the right hand limit of the real portion. – Doug M Jun 2 '16 at 1:55
• hint: consider the sequences $z_n=1/n$ and $\tilde z_n=-1/n$, $n\in\mathbb N$. – sranthrop Jun 2 '16 at 1:55
• Just looking at the reals, it becomes unbounded when $z > 0$. – Jared Jun 2 '16 at 1:56
• @Jared, but one has the stronger result that the limit doesn't exist even in the one-point compactification (whereas $\lim_{\substack{x \in \mathbb R \\ x \to 0^+}} e^{1/x}$ does). – LSpice Jun 2 '16 at 2:02

if the limit exists then any sequence tending to zero must give the same answer in the limit.

So consider $x_n = 1/n$ and $y_n = -1/n.$ One gives infinity, the other gives zero.

done.

• Sorry could you explain to me why sequences are necessary here? Also why did you choose $1/n$ and $-1/n$? Thanks for helping :) – Faolan Jun 2 '16 at 2:03
• It might be helpful if you could show me if you don't mind. I'm a little confused sorry – Faolan Jun 2 '16 at 2:04
• @Faolan Why can't you try $1/n,-1/n$ yourself? Plug it in, and observe ... – zhw. Jun 2 '16 at 2:21
• So you just plug them into $e^(1/z)$? – Faolan Jun 2 '16 at 2:23
• it's a theorem that the limit exists if and only if the values along every sequence converge and they converge to the same limit,. – Mark Joshi Jun 2 '16 at 4:28

Value of e is fixed i.e. $e=2.718$ So if $\lim_{z \to 0}$ then $e^{\frac{1}{z}}$ is tending to infinity or we can say $e^{\frac{1}{z}}$ does not exist .