Question

$\underset{x\to 0^{+}}{lim}x(e^{-x}+e^{-2x}+e^{-3x}+....\infty )$ is equal to =

• A. 0
• B. -1
• C. 1
• D. does not exits

Answer 1

Answer: (C)

1

The series $(e^{-x}+e^{-2x}+e^{-3x}+.........\infty )$ is an infinite G.P. in which,

first term, $a=e^{-x}$

common ratio, $r=e^{-x}$

$S_{\infty }=\frac{a}{1-r}$

$=\frac{e^{-x}}{1-e^{-x}}$

$=\frac{1}{\frac{1}{e^{-x}}-\frac{e^{-x}}{e^{-x}}}$

$=\frac{1}{e^x-1}$

Now,

$\underset{x\to 0^{+}}{lim}x(e^{-x}+e^{-2x}+e^{-3x}+.........\infty )$

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

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