The limit of left[frac{1}{x^{2}}+frac{(2013)^ | vedclass

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Question

(A) We are evaluating $\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{e^{x}-1}\right\}$.Using the standard limit $\lim_{x \rightarr

Vedclass · Accepted Answer

approaches $+\infty$

Vedclass · Answer

(A) We are evaluating $\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{e^{x}-1}\right\}$.Using the standard limit $\lim_{x \rightarrow 0} \frac{e^x-1}{x} = 1$,we can rewrite the expression as:$\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{x} \cdot \frac{x}{e^{x}-1}\right\}$.We know that $\lim _{x \rightarrow 0} \frac{(2013)^{x}-1}{x} = \ln(2013)$ and $\lim _{x \rightarrow 0} \frac{x}{e^{x}-1} = 1$.Thus,the expression becomes $\lim _{x \rightarrow 0} \frac{1}{x^{2}} + \ln(2013) \cdot 1$.As $x \rightarrow 0$,$\frac{1}{x^2} \rightarrow +\infty$.Therefore,the limit is $+\infty$.

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

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