$\lim _{x\rightarrow 0^{-}}e^\left(1/x\right)=0$ ($\delta-\varepsilon$ definition)

Question

$\lim _{x\rightarrow 0^{-}}e^\left(1/x\right)=0$ So I wrote the definition: $\forall \varepsilon>0, \exists \delta_{\varepsilon}$ such that if $x_{0}-\delta<x<x_{0}$ then $\left|f(x)-L\right|<\varepsilon$ So let $\varepsilon>0$ be given. Choose $\delta=-\dfrac{1}{\ln(\varepsilon)}$ if $-\delta<x<0$ implies that $\left|e^\left(1/x\right)\right|<\varepsilon$. So did I make a mistake? Because when I take $\varepsilon=0.001$ I didn't get closer to $0$ $\varepsilon=0.001$ then $\delta=0.144$ then the interval is $0.144<x<0$. So for example if I take $0.140$ I can't even get closer to $0$. For instance: $e^\left(1/0.140\right)=1265.03...$

Answer 1

Your mistake was that you took $0\lt x\lt \delta$, when you took $x=0.140$.

You should take $-\delta\lt x\lt 0$, and then your requirement will be satisfied.

E.g., $e^{1/-0.140}\approx0.00079<0.001$.

Answer 2

If $\varepsilon\geqslant1$, any $\delta$ will do. Otherwise, you have\begin{align}\left|e^{1/x}\right|<\varepsilon&\iff e^{1/x}<\varepsilon\\&\iff\frac1x<\ln\varepsilon\\&\iff x>\frac1{\ln\varepsilon}.\end{align}So, take $\delta=-\frac1{\ln\varepsilon.}$

Answer 3

Option:

Consider $y:=-x$ and limit $y \rightarrow 0^{+}$.

$|e^{-1/y}|=|\dfrac{1}{e^{1/y}}|=|\dfrac{1}{1+1/y....}|\le |y|.$

Choose $\delta=\epsilon.$