Question

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

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

Answer 1

The correct option is B $\frac{3}{2}$

Given Limit $=e^{\underset{x\to \infty }{lim}(\frac{a}{x}-\frac{4}{x^2})\times 2x}$
$=e^{\underset{x\to \infty }{lim}(2a-\frac{8}{x})}$
$=e^{2a}$
And this is equal to $e^3$
On equating we get,
$\therefore 2a=3$
Hence $a=\frac{3}{2}$