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  4. The limit of [(1/x2)+((2013)x-1/ex-1)×(1/ex-1)] as x → 0

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Q. 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 $x → 0$

WBJEEWBJEE 2013Limits and Derivatives

Solution:

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