Question

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

• 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 $+\infty$

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

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

$\underset{x\to 0}{lim}[\frac{1}{x^2}+\frac{{\left(2013\right)}^x-1}{x}.\frac{x}{e^x-1}]$

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

$=+\infty +\log \left(2013\right).1$

$=+\infty$