Question

The limit of $[\frac{1}{x^2}+\frac{{\left(2013\right)}^x}{e^x-1}-\frac{1}{e^x-1}]asx\to 0$

• A. Approaches +$\infty$
• B. Approaches -$\infty$
• C. Is equal to ${log}_e\left(2013\right)$
• D. Does not exist

Answer 1

The correct option is A Approaches +$\infty$

Given,

${lim}_{x\to 0}[\frac{1}{x^2}+\frac{{2013}^x}{e^x-1}-\frac{1}{e^x-1}]$

$={lim}_{x\to 0}\left(\frac{1}{x^2}\right)+{lim}_{x\to 0}\left(\frac{{2013}^x}{e^x-1}\right)-{lim}_{x\to 0}\left(\frac{1}{e^x-1}\right)$

$=\infty +[{lim}_{x\to 0+}\left(\frac{{2013}^x}{e^x-1}\right)=\infty ,{lim}_{x\to 0-}\left(\frac{{2013}^x}{e^x-1}\right)=-\infty ]-$

$[{lim}_{x\to 0+}\left(\frac{1}{e^x-1}\right)=\infty ,{lim}_{x\to 0-}\left(\frac{1}{e^x-1}\right)=-\infty ]$

$=\infty +\left[diverges\right]-\left[diverges\right]$

$=\infty$