Continuity and limits with transcendental functions: Determine the interval on which following functions

$f\left(x\right)=\frac{{e}^{x}}{1-{e}^{x}}$

Kye
2021-10-31
Answered

Continuity and limits with transcendental functions: Determine the interval on which following functions

$f\left(x\right)=\frac{{e}^{x}}{1-{e}^{x}}$

You can still ask an expert for help

lobeflepnoumni

Answered 2021-11-01
Author has **99** answers

Given

$f\left(x\right)=\frac{{e}^{x}}{1-{e}^{x}}$

f(x) is continuous for all points except for the point where denominator is equal to 0.

$1-{e}^{x}=0$

${e}^{x}=1$

${\mathrm{ln}e}^{x}=\mathrm{ln}1$

$x=0$

Thus interval of continuity are

$(-\mathrm{\infty},0)$ and $(0,\mathrm{\infty})$

$\underset{x\to {0}^{-}}{lim}f\left(x\right)=\underset{x\to {0}^{-}}{lim}\frac{{e}^{x}}{1-{e}^{x}}$

$=\underset{h\to 0}{lim}\frac{{e}^{-h}}{1-{e}^{-h}}$

$=\underset{h\to 0}{lim}\frac{1}{\left(\frac{{e}^{h}-1}{h}\right)h}$

$=\underset{h\to 0}{lim}\frac{1}{h}$

$=\mathrm{\infty}$

$\underset{x\to {0}^{+}}{lim}f\left(x\right)=\underset{x\to {0}^{+}}{lim}\frac{{e}^{x}}{1-{e}^{x}}$

$=\underset{h\to 0}{lim}\frac{{e}^{h}}{-({e}^{h}-1)}$

$=\underset{h\to 0}{lim}\frac{{e}^{h}}{-\left(\frac{{e}^{h}-1}{h}\right)h}$

$=\underset{h\to 0}{lim}(-\frac{{e}^{h}}{h})$

$=-\mathrm{\infty}$

f(x) is continuous for all points except for the point where denominator is equal to 0.

Thus interval of continuity are

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