Question

If $\underset{x\to \infty }{\lim }{\left(1+\frac{a}{x}-\frac{4}{x^2}\right)}^{2x}=e^3$ , then ${}^{\prime }{a}^{\prime }$ is equal to :

• A. $2$
• B. $\frac{3}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$Lim{\left(1+\frac{a}{2}-\frac{4}{x^2}\right)}^{2x}\left(1\infty form\right)$
$=e^{\left[{\lim }_{x\to \infty }\left(1+\frac{a}{x}\frac{4}{x^2}-1\right)2x\right]}$
$=e\underset{x\to \infty }{\lim }$$\left(2a-\frac{8}{x}\right)=e^{2a}$
$\therefore 2a=3\Rightarrow a=3/2$