Question

If $\underset{x\to \infty }{\lim }{\left(1+\frac{\lambda }{x}+\frac{\mu }{x^2}\right)}^{2x}=e^2$ then

• A. $\lambda =-1,\mu =2$
• B. $\lambda =2,\mu =1$
• C. $\lambda =1,\mu =$ any real number
• D. $\lambda =\mu =$ any real number

Answer 1

Answer: (C)

$\underset{x\to \infty }{\lim }{\left(1+\frac{\lambda }{x}+\frac{\mu }{x^2}\right)}^{2x}$ [$1^{\infty }$ form]
$=e^{\underset{x\to \infty }{\lim }\left(1+\frac{\lambda }{x}+\frac{\mu }{x^2}-1\right)2x}=e^{\underset{x\to \infty }{\lim }\left(2\lambda +\frac{\mu }{x^2}\right)}=e^{2\lambda }$ for any value of $\mu$
Given $e^{2\lambda }=e^2\Rightarrow \lambda =1\therefore \lambda =1$ and $\mu$ is any real number.