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WBJEE · Maths · Limits

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

  1. A approaches \(+\infty\)
  2. B approaches \(-\infty\)
  3. C is equal to \(\log_{e}\), (2013)
  4. D does not exist
Verified Solution

Answer & Solution

Correct Answer

(A) approaches \(+\infty\)

Step-by-step Solution

Detailed explanation

\(\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right\}\) \(=\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{e^{x}-1}\right\}\)…