Q. The value of $\lim_{x \rightarrow \infty}\left( \frac{x + 3}{x + 1} \right)^{x + 2}$ is

A) 0
B) 1
C) $e^{2}$
D) $e^{4}$
Explanation

(c)
$\lim_{x \rightarrow \infty}\left( \frac{x + 3}{x + 1} \right)^{x + 2} = \lim_{x \rightarrow \infty}\left\lbrack \left( 1 + \frac{2}{x + 1} \right)^{\left( \frac{x + 2}{2} \right)} \right\rbrack^{\frac{2}{x + 2} \times (x + 2)}$
$= \lim_{x \rightarrow \infty}\left( 1 + \frac{2}{x + 1} \right)^{\left( \frac{x + 2}{2} \right)\left\lbrack \frac{2(x + 2)}{x + 1} \right\rbrack}$
$= e^{\lim_{x \rightarrow \infty}\frac{(2 + 4/x)}{(1 + 1/x)}}$
$= e^{2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.