0
$\begingroup$

I want to show that the extended exponential function $\text{Exp}:\mathbb R \cup \{\pm \infty\} \to \mathbb R \cup \{\pm \infty\}$, $$\text{Exp($x$)} = \begin{cases}\exp(x) & x\in \mathbb R\\0 & x = -\infty \\ +\infty & x= +\infty \end{cases} $$ is continuous. I know that $\exp(x)$ is continuous on $\mathbb R$. I could show it pretty easily using the sequence criterion for continuity: Let $(a_n)_{n\in \mathbb N}$ a sequence in $\mathbb R$ with $\lim_{n\to \infty} a_n = +\infty$. It easily follows that, since $\exp(x)$ is continuous, that $$\lim_{n\to \infty} \exp(a_n) = \exp(\lim_{n\to\infty} a_n) = \exp(+\infty) = +\infty$$ (analogous for $x = - \infty$). However, I tried it directly with the $\epsilon-\delta$-criterion too but couldn't get far since I don't know how for example the expression $|x-(-\infty)|<\delta$ is to be interpreted. Could someone give me a proof or an ansatz for it?

  • 3
    If you want to use the $\varepsilon$-$\delta$ criterion, you first need a metric on $[-\infty, +\infty]$, and one that induces the standard topology. One such metric is $d(x,y) = \lvert \arctan x - \arctan y\rvert$, where $\arctan (+\infty) = \frac{\pi}{2}$ and $\arctan (-\infty) = -\frac{\pi}{2}$. Another is $d(x,y) = \bigl\lvert \frac{x}{1 + \lvert x\rvert} - \frac{y}{1 + \lvert y\rvert}\bigr\rvert$, with $\pm\infty$ mapped to $\pm 1$ under $x\mapsto \frac{x}{1+\lvert x\rvert}$. – 2017-01-09
  • 0
    Okay that seems far too complicated. Is my solution using the sequences correct? – 2017-01-09
  • 1
    You can only use the continuity of $\exp$ to move a limit inside if $\lim a_n \in \mathbb{R}$. To show continuity of $\operatorname{Exp}$ at $\pm\infty$, use the monotonicity of $\exp$. If $a_n \to +\infty$, you want to show $\exp(a_n) \to +\infty$. So for every $K \in \mathbb{R}$, you need an $N$ such that $\exp(a_n) \geqslant K$ for $n \geqslant N$. As $\exp (x) \geqslant K \iff x \geqslant \log K$ [we're only considering $K > 0$, $\exp(x)\geqslant K$ is trivially true for $K\leqslant 0$], you need an $N$ with $a_n \geqslant \log K$ for $n\geqslant N$. That follows from $a_n\to +\infty$. – 2017-01-09

0 Answers 0