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The limit of $\left[\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right]$ as $x \rightarrow 0$
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approaches $+\infty$
$\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{e^{x}-1}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{x} \cdot \frac{x}{e^{x}-1}\right\}$
$=\lim _{x \rightarrow 0} \frac{1}{x^{2}}+\lim _{x \rightarrow 0} \frac{(2013)^{x}-1}{x} \cdot \lim _{x \rightarrow 0} \frac{x}{e^{x}-1}$
$=+\infty+\log (2013) \cdot 1$
$=+\infty$
$=\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{e^{x}-1}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{x} \cdot \frac{x}{e^{x}-1}\right\}$
$=\lim _{x \rightarrow 0} \frac{1}{x^{2}}+\lim _{x \rightarrow 0} \frac{(2013)^{x}-1}{x} \cdot \lim _{x \rightarrow 0} \frac{x}{e^{x}-1}$
$=+\infty+\log (2013) \cdot 1$
$=+\infty$
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