Question

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

• A. $e^{4}$
• B. $0$
• C. 1
• D. $e^{2}$

Answer 1

Answer: (D)

$e^{2}$

Answer 2

$\lim_{x \rightarrow \infty}\left( \frac{x + 3}{x + 1} \right)^{x + 2} = \lim_{x \rightarrow \infty}\left( 1 + \frac{2}{x + 1} \right)^{\frac{x + 1}{2}.(x + 2).\frac{2}{(x + 1)}}$

$= \lim_{x \rightarrow \infty}\left( \left( 1 + \frac{2}{x + 1} \right)^{\frac{x + 1}{2}} \right)^{2.\left( \frac{1 + \frac{2}{x}}{1 + \frac{1}{x}} \right)} = e^{2\lim_{x \rightarrow \infty}\left\lbrack \frac{\left( 1 + \frac{2}{x} \right)}{\left( 1 + \frac{1}{x} \right)} \right\rbrack} = e^{2}.$

Alternative method : $\lim_{x \rightarrow \infty}\left( \frac{x + 3}{x + 1} \right)^{x + 2}$

=$\lim_{x \rightarrow \infty}\left( 1 + \frac{2}{x + 1} \right)^{x + 2} = e^{\lim_{x \rightarrow \infty}\frac{2}{x + 1}(x + 2)} = e^{\lim_{x \rightarrow \infty}2\left( \frac{1 + \frac{2}{x}}{1 + \frac{1}{x}} \right)} = e^{2}$