better way to explain $\lim\limits_{x\to 0} e^ \frac {−1}{|x|} $

Question

$$\lim\limits_{x\to 0} e^\frac{−1}{|x|}$$

I know that if $x$ goes to $0$, that $\frac{-1}{|x|}$ goes to $-\infty$ and thus, the limit goes to $0$.

But, is there a more mathematical way to explain it rather than with words?

Edited the formatting, want to make sure this is what you meant — WaveX, Oct 14 '18 at 12:28
Maybe state it as a continuity property of a composition of $\frac{1}{|x|}$ and $e^x$. — Henno Brandsma, Oct 14 '18 at 12:34
As a part of basic properties of $e^x$ one learns that $e^x\to\infty$ as $x\to\infty$. Your limit is an easy consequence of that property and the standard limit $\lim_{x\to\infty} (1/x)=0$. In essence what you have said in words in your post is the correct justification. — Paramanand Singh, Oct 14 '18 at 21:01

Calvin Khor · Answer 1 · 2018-10-14T12:34:37.067

0

One way to proceed is via $$ e^{-x} \le \frac1{1+x}$$ which itself follows because $ 1+x \le e^x$ for all $x\ge 0$. Then

$$0\le e^{-1/|x|} \le \frac1{1+1/|x|} = \frac{|x|}{|x|+1} \le |x| $$ and $|x|\to 0$ as $x\to 0$, so we can conclude with the "squeeze rule".

edited Oct 14 '18 at 12:34

answered Oct 14 '18 at 12:28

Calvin Khor

score 0 · Answer 2 · answered Oct 14 '18 at 14:09

Use the definition. Take $\varepsilon \in (0 , \infty)$ and distinguish two cases:

if $\varepsilon \in [1 , \infty)$, then $\log \varepsilon \in [0 , \infty)$ and each $\delta \in (0 , \infty)$ is valid: for all $\varepsilon \in [1 , \infty)$, we have (take some $\delta \in (0 , \infty)$), $$ - \frac{1}{|x|} < \log \varepsilon \quad \Longrightarrow \quad e^{- \frac{1}{|x|}} < \varepsilon $$ for $x \in (- \delta , \delta) \setminus \{0\}$.
if $\varepsilon \in (0 , 1)$, then take $\delta = - \frac{1}{\log \varepsilon}$: for $x \in (- \delta , \delta) \setminus \{0\}$, we have $$ |x| < - \frac{1}{\log \varepsilon} \quad \Longrightarrow \quad - \frac{1}{|x|} < \log \varepsilon \quad \Longrightarrow \quad e^{- \frac{1}{|x|}} < \varepsilon\mbox{.} $$

2 Answers2