Question

The value of $\underset{y\to \infty }{lim}{\left(\begin{array}{c}1+\frac{x}{y}\end{array}\right)}^{2y}$ is

• A. $e^{2x}$
• B. $e^{-2x}$
• C. $e^x$
• D. $e^{-x}$

Answer 1

The correct option is A $e^{2x}$

$\underset{y\to \infty }{lim}{(l+\frac{x}{y})}^{2y}=z$

Now, when limit are applied, it is of indeterminate form $1^{\infty }$, so, writing it after applying $\log$ on both sides,

$\underset{y\to \infty }{lim}2y\log (1+\frac{x}{y})=\log z$

Now, $LHS$ is of indeterminate form $0\times \infty$ as $\underset{y\to \infty }{lim}2y=\infty$ and

$\underset{y\to \infty }{lim}\log (1+\frac{x}{y})=0$

So, Writing it as $\frac{f\left(x\right)}{\varphi \left(x\right)}$ where, $f\left(x\right)=\log (1+\frac{x}{y})$ and $\varphi \left(x\right)=2y$

$\underset{y\to \infty }{lim}\frac{\log (1+\frac{x}{y})}{\frac{1}{2y}}=\log yz$

Now, it is of indeterminate $\frac{0}{0}$, So applying L Hospital Rule,

$\underset{y\to \infty }{lim}\frac{\frac{1}{(1+\frac{x}{y})}\left(\frac{2x}{y^2}\right)}{\left(\frac{1}{2y^2}\right)}=\log yz$

$\underset{y\to \infty }{lim}\frac{2x}{(1+\frac{x}{y})}=\log yz$

Now applying limit $\frac{2x}{(1+0)}=\log yz$

$2x=\log yz$

$zy=e^{2x}$

Hence $e^{2x}$ is the correct option