Question

$\lim _{x \rightarrow \infty}\left(1-\frac{4}{x-1}\right)^{3 x-1}$ is equal to

(A) $e^{12}$

(B) $e^{-12}$

(C) $e^4$

(D) $e^3$

Answer 1

$L=\lim _{x \rightarrow \infty}\left[\left(1-\frac{4}{x-1}\right)^{\frac{-(x-1)}{4}}\right]^{-4\left(\frac{3 x-1}{x-1}\right)}$

$=e^{\left.-4 \lim _{x \rightarrow-}\left(3-\frac{1}{x}\right)\right)\left(1 \frac{1}{x}\right)}$

$=e^{-4 \times 3}=e^{-12}$

So, The correct option will be (B).