Question

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

• A. $\mu =2,\lambda =$ any Real Constant
• B. $\lambda =2,\mu =1$
• C. $\lambda =1,\mu =$ any Real Constant
• D. $\lambda =\mu =2$

Answer 1

Answer: (C)

$\lambda =1,\mu =$ any Real Constant

Since the limit is of the form, $1^{\infty }$........(substitute the value of x)

We rewrite the expression in the form of,

$\underset{x\to \infty }{lim}\exp \left[2x(\frac{\lambda }{x}+\frac{\mu }{x^2})\right]$

$\Rightarrow \underset{x\to \infty }{lim}\exp [2\lambda +\frac{\mu }{x}]$
Since, this has to be equal to $e^2$
$\lambda =1$
Whereas $\mu$ can be any real number.

Hence, option C.