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Q. If $\displaystyle\lim_{x\to\infty} \left(1+\frac{\lambda}{x}+\frac{\mu}{x^{2}}\right)^{2x}\quad = e^{2} $ then

Limits and Derivatives

Solution:

$\displaystyle\lim_{x\to\infty} \left(1+\frac{\lambda}{x}+\frac{\mu}{x^{2}}\right)^{2x}\quad$ [$1^{\infty}$ form]
$= e^{\displaystyle\lim _{x\to \infty } \left(1+\frac{\lambda }{x}+\frac{\mu }{x^{2}}-1\right)2x} = e^{\displaystyle\lim _{x\to \infty } \left(2\lambda+\frac{\mu }{x^{2}}\right)} = e^{2\lambda}$ for any value of $\mu$
Given $e^{2\lambda } = e^{2} \Rightarrow \lambda = 1 \quad \therefore \lambda = 1$ and $\mu$ is any real number.