Question

$\displaystyle\lim_{x \to \infty} \left( \frac{x+6}{x+1}\right)^{x+4} =$

• A. $e^3$
• B. $e^5$
• C. $e^4$
• D. Does not exist

Answer 1

Answer: (B)

$e^5$

Answer 2

We have,
$\displaystyle\lim_{x\to\infty}\left(\frac{x+6}{x+1}\right)^{x+4} = \displaystyle\lim_{x\to\infty}\left(\frac{1+\frac{6}{x}}{1+\frac{1}{x}}\right)^{x+4}$
$=\displaystyle\lim_{x\to\infty} \frac{\left(1+\frac{6}{x}\right)^{x}}{\left(1+\frac{1}{x}\right)^{x}} \frac{\left(1+\frac{6}{x}\right)^{4}}{\left(1+ \frac{1}{x}\right)^{4}} =\frac{e^{6}}{e} =e^{5}$