ExamBro
ExamBro
WBJEE · Maths · Limits

The \(\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}\) equals

  1. A 1
  2. B 0
  3. C \(\mathrm{e}^{-8 / 3}\)
  4. D \(\mathrm{e}^{-4 / 9}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{e}^{-8 / 3}\)

Step-by-step Solution

Detailed explanation

\(\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}=1^{\infty}=e^{L}=e^{-\frac{8}{3}}\) \(L=\lim _{x \rightarrow \infty} 4 x\left(\frac{3 x-1}{3 x+1}-1\right)=\lim _{x \rightarrow \infty} 4 x\left(\frac{-2}{3 x+1}\right)=\frac{-8}{3}\)