Question

$\underset{x\to \infty }{lim}(1+\frac{\lambda }{x}+\frac{\mu }{x^2}{)}^{2x}=e^4$ then

• A. $\lambda =\mu =1$
• B. $\lambda =2,\mu =1$
• C. $\lambda =1,\mu =2$
• D. $\lambda =1,\mu =0$

Answer 1

The correct option is B $\lambda =2,\mu =1$

$\underset{x\to \infty }{lim}(1+\frac{\lambda }{x}+\frac{\mu }{x^2}{)}^{2x}=e^4$
$\Rightarrow \underset{x\to \infty }{lim}{(1+\frac{\lambda x+\mu }{x^2})}^{2x}=e^4$

$Now,\underset{x\to \infty }{lim}f(x{)}^{g\left(x\right)}=e^{(\underset{x\to \infty }{lim}\left(f\right(x)-1).g\left(x\right))}$

Using above formula we get,

$\Rightarrow e^{\underset{x\to \infty }{lim}\left(\frac{\lambda x+\mu }{x^2}\right)\times 2x}=e^4$

$\Rightarrow e^{2\underset{x\to \infty }{lim}\left(\frac{\lambda x+\mu }{x}\right)}=e^4$
$\Rightarrow e^{2\underset{x\to \infty }{lim}(\lambda +\frac{\mu }{x})}=e^4$
$\Rightarrow e^{2\lambda }=e^4\Rightarrow \lambda =2,\mu \in R$