TS EAMCET · Maths · Limits
\(\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}\) is equal to
- A \(e^4\)
- B \(e^6\)
- C \(e^5\)
- D \(e\)
Answer & Solution
Correct Answer
(C) \(e^5\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}=\lim _{x \rightarrow \infty}\left(1+\frac{5}{x+1}\right)^{x+4} \\ & =e^{5 \lim _{x \rightarrow \infty}\left(\frac{x+4}{x+1}\right)}=x \\ & =e^5\end{aligned}\)
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