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

none of these

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

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

Whereas $\mu$ can be any real number.

Hence (4) is the correct answer.