Question

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

Answer 1

$\underset{x\to \infty }{lim}(1+\frac{a}{x}-\frac{4}{x^2}{)}^{2x}=e^{\underset{x\to \infty }{lim}(1+\frac{a}{x}-\frac{4}{x^2}-1)2x}=e^{2a}$

$⟹e^3=e^{2a}⟹a=\frac{3}{2}$