If limlimits_x → ∞≤ft(1+(λ)/x+(μ)/(x^2))^2 x=e^2, then

$\lim\limits_{x \rightarrow \infty}\left(1+\frac{1}{\frac{x^2}{\lambda x+\mu}}\right)^{2 x}=\lim\limits_{x \rightarrow \infty}\left(1+\frac{1}{\frac{x^2}{\lambda x+\mu}}\right)^{\frac{x^2}{\lambda x+\mu} \times \frac{2(\lambda x+\mu)}{x}}=e^{\lim\limits_{\lambda \rightarrow \infty}\left(2 \lambda+\frac{2 \mu}{x}\right)}$

$e^{2 \lambda}=e^2 \Rightarrow \lambda=1$

Whereas $\mu$ can be any real number.

Hence (4) is the correct answer.