Question

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

• A. approaches $+\infty$
• B. approaches $-\infty$
• C. is equal to $lo{g}_e$ (2013)
• D. does not exist

Answer 1

Answer: (A)

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