WBJEE · Maths · Limits
The limit of \(\left[\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right]\) as \(x \rightarrow 0\)
- A approaches \(+\infty\)
- B approaches \(-\infty\)
- C is equal to \(\log_{e}\), (2013)
- D does not exist
Answer & Solution
Correct Answer
(A) approaches \(+\infty\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right\}\) \(=\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{e^{x}-1}\right\}\)…
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