Question

Show that $\underset{x\to 0}{\lim }{e}^{-1/x}$ does not exist.

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }{e}^{-\frac{1}{x}} \\ \mathrm{Left}\mathrm{hand}\mathrm{limit}: \\ \underset{x\to 0^{-}}{\lim }\left(e^{-\frac{1}{x}}\right) \\ \mathrm{Let}x=0-h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }{e}^{\left(\frac{-1}{0-h}\right)} \\ =\underset{h\to 0}{\lim }\left(e^{\frac{1}{h}}\right) \\ =\underset{h\to 0}{\lim }\left(e^{\frac{1}{h}}\right)\left[\mathrm{When}h\to 0,\mathrm{then}\frac{1}{h}\to \infty .\right] \\ =e^{\infty } \\ =\infty \\ \mathrm{Right}\mathrm{hand}\mathrm{limit}: \\ \underset{x\to 0^{+}}{\lim }\left(e^{-\frac{1}{x}}\right) \\ \mathrm{Let}x=0+h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left(e^{\frac{-1}{0+h}}\right) \\ =\underset{h\to 0}{\lim }\left(\frac{1}{e^{{\displaystyle \frac{1}{h}}}}\right) \\ =\frac{1}{\infty } \\ =0 \\ \underset{x\to 0^{-}}{\lim }{e}^{-\frac{1}{x}}\ne \underset{x\to 0^{+}}{\lim }{e}^{\frac{1}{x}} \\ \mathrm{Thus},\underset{x\to 0}{\mathrm{limit}}{e}^{-\frac{1}{x}}\mathrm{does}\mathrm{not}\mathrm{exist}. \end{gathered}$$