Question

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

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

Answer 1

Answer: (A), (C), (D)

$\lambda =1,\mu$= any real constant
$\lambda =\mu =1$
$\lambda =1,\mu =2$

$\underset{x\to \infty }{lim}{(1+\frac{\lambda }{x}+\frac{\mu }{x^2})}^{2x}=e^2\Rightarrow \underset{x\to \infty }{lim}{\left({(1+\frac{\lambda }{x}+\frac{\mu }{x^2})}^{1/\frac{\lambda }{x}+\frac{\mu }{x^2}}\right)}^{2x(\frac{\lambda }{x}+\frac{\mu }{x^2})}=e^2\Rightarrow e^{\underset{x\to \infty }{lim}2(\lambda +\frac{\mu }{x})}=e^2\Rightarrow e^{2\lambda }=e^2$
$\Rightarrow \lambda =1$ and $\mu$ is any real constant