WBJEE · Maths · Limits
The \(\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}\) equals
- A 1
- B 0
- C \(\mathrm{e}^{-8 / 3}\)
- D \(\mathrm{e}^{-4 / 9}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{e}^{-8 / 3}\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}=1^{\infty}=e^{L}=e^{-\frac{8}{3}}\) \(L=\lim _{x \rightarrow \infty} 4 x\left(\frac{3 x-1}{3 x+1}-1\right)=\lim _{x \rightarrow \infty} 4 x\left(\frac{-2}{3 x+1}\right)=\frac{-8}{3}\)
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