$\lim\limits_{x→∞}\left(\frac{x-7}{x+1}\right)^x$ is equal to :

$e^{-8}$

The correct answer is Option (3) → $e^{-8}$

$y=\lim\limits_{x→∞}\left(\frac{x-7}{x+1}\right)^x$

$y=\lim\limits_{x→∞}(1-\frac{8}{x+1})^x$

$\log y=\lim\limits_{x→∞}x\log(1-\frac{8}{(x+1)})$

$=\frac{\log(1-\frac{8}{(x+1)})}{x^{-1}}$

$\log y=\lim\limits_{x→∞}\frac{\log(1-\frac{8}{x+1})}{\frac{1}{x}}$ Using L'Hopital's rule

$\log y=\lim\limits_{x→∞}\frac{(1-\frac{8}{(x+1)})^{-1}(\frac{8}{(x+1)^2})}{\frac{-1}{x^2}}$

$=\lim\limits_{x→∞}-\frac{x^2}{(x+1)^2}\frac{(8)}{(1-\frac{8}{(x+1)})}$

$=\lim\limits_{x→∞}-\frac{1}{(1+\frac{1}{x})^2}\frac{(8)}{(1-\frac{8}{(x+1)})}$

$\log y=-8$

$⇒y=e^{-8}$