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Find
$$L=\lim_{z \rightarrow 0} e^{z}.$$

First I simply wrote down the answer $1$ but I realise that I can't just mimic things I'd do in the complex case that I would have done for the real case. I then let $z = x+iy$ where $x,y$ are real so the limit is
$$L = \lim_{(x,y) \rightarrow (0,0)} e^x e^{iy}\\ = \lim_{(x,y) \rightarrow 0} |e^x e^{iy}| = \lim_{(x,y)\rightarrow (0,0)} e^x$$
which does in fact equal $1$ (just a simple limit in the real case).
I was wondering - did I really have to go through these steps (transforming it into a multivariable limit)? How would I justify the limit?

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  • $\begingroup$ As $\exp(z)$ is holomorphic it is also continuous, hence the limit is simply $\exp(0)$. But the solution you gave is more basic, which is probably what you are asked to do in the exercise. $\endgroup$ – MrYouMath Jul 22 '17 at 12:27
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    $\begingroup$ In both arguments, at the very last step you did exactly the same: invoking the continuity of $e^z$ at $z=0$. If you added $\lim_{(x,y)\to(0,0)}e^{x}=\lim_{x\to0}e^x$ then it would be clear that you are using the continuity of $e^x$ at $x=0$ as a real function. $\endgroup$ – Hellen Jul 22 '17 at 12:28
  • $\begingroup$ @Hellen: But in the last line he only had a univariate limit, which means he used $\exp(x)$ is continuous. Is it really necessary to show that $\exp(x)$ is continuous in $x\in \mathbb{R}$? $\endgroup$ – MrYouMath Jul 22 '17 at 12:30
  • $\begingroup$ @MrYouMath No. He still has a function of complex variable $z\mapsto e^{Re(z)}$ $\endgroup$ – Hellen Jul 22 '17 at 12:31
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    $\begingroup$ @Twenty-sixcolours well, we usually do the same in multivariable limits too. Wether are you in the complex or in the real world, we first check continuity - if our function is continuous, nothing needs to be done and the limit is simply evaluating the function. The problem is where continuity fails - and that's where calculating limits gets interesting, and all your techniques kicks in. $\endgroup$ – Henrique Augusto Souza Jul 22 '17 at 17:07
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Since$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots,$$you know that$$0\leqslant|e^z-1|=\left|z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right|\leqslant|z|+\frac{|z|^2}{2!}+\frac{|z|^3}{3!}+\cdots=e^{|z|}-1.$$Since $\lim_{z\to0}e^{|z|}-1=0$, it follows that $\lim_{z\to0}|e^z-1|=0$, which is the same thing as stating that $\lim_{z\to0}e^z-1=0$.

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