Question

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

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

Answer 1

The correct option is A Approaches $+\infty$

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

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

$=\left(\underset{x\to 0}{lim}\frac{1}{x^2}\right)+\left(\log 2013\right)\times 1\left[\underset{x\to 0}{lim}\frac{a^x-1}{x}={\log }_{e}a;\underset{x\to 0}{lim}\frac{e^x-1}{x}=1\right]$

$=\infty +\log 2013=\infty$

as $x\to 0\Rightarrow$Limit $\to \infty$