Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

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

$={\perp }^{\infty }$ form.

So. if $\underset{x\to \infty }{lim}\left[f\right(x){]}^{g\left(x\right)}={\perp }^{\infty }$ form

thus $\underset{e^{x\to \infty }}{lim}\left[f\right(n)-1]g\left(x\right)$

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

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

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

$e^{2\lambda +0}=e^4\Rightarrow 2\lambda =4\Rightarrow$

$\Rightarrow \lambda =2$

$\Rightarrow \lambda =\mu =2$