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?

Answer 1

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".

Answer 2

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 $$ |x| < - \frac{1}{\log \varepsilon} \quad \Longrightarrow \quad - \frac{1}{|x|} < \log \varepsilon \quad \Longrightarrow \quad e^{- \frac{1}{|x|}} < \varepsilon\mbox{.} $$