AP EAMCET · Maths · Limits
For \(A \neq 0, x < 0, \lim _{n \rightarrow \infty} \frac{\sin x-e^{n x}}{1+A e^{n x}}=\)
- A \(\frac{1}{A}\)
- B \(\sin x\)
- C \(-\frac{1}{A}\)
- D \(-\sin x\)
Answer & Solution
Correct Answer
(C) \(-\frac{1}{A}\)
Step-by-step Solution
Detailed explanation
Given, \(\lim _{n \rightarrow \infty} \frac{\sin x-e^{n x}}{1+A e^{n x}}=\lim _{n \rightarrow \infty} \frac{e^{n x}\left(\frac{\sin x}{e^{n x}}-1\right)}{e^{n x}\left(\frac{1}{e^{n x}}+A\right)}\)…
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