Q91 / 155 · WBJEE 2021 · Mathematics

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

  • A
    1
  • B
    0
  • Correct answer
    C
    $\mathrm{e}^{-8 / 3}$
  • D
    $\mathrm{e}^{-4 / 9}$

Solution

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

Hence, the correct answer is option: 'C'.

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