Let, $y=\underset{n\to \infty }{\lim }P{\left(1+\frac{r}{100n}\right)}^{tn}$

Let, $n=\frac{1}{x}$ so when $n\to \infty$ then $x\to 0$.

$\therefore y= \underset{x\to 0}{\lim }P{\left(1+\frac{rx}{100}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}$

Taking log on both sides, we get

$\ln y= \underset{x\to 0}{\lim } \ln P{\left(1+\frac{rx}{100}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}$

$\Rightarrow \ln y= \underset{x\to 0}{\lim } \left[\ln P+\frac{t}{x}\ln \left(1+\frac{rx}{100}\right)\right]$

$\Rightarrow \ln y=\ln P+\underset{x\to 0}{\lim } t\frac{\ln \left(1+\frac{rx}{100}\right)}{{\displaystyle \frac{rx}{100}}}\times \frac{r}{100}$

$\Rightarrow \ln y-\ln P=\frac{rt}{100}\left[\because \underset{x\to 0}{\lim }\frac{\ln \left(1+x\right)}{x}=1\right]$

$\Rightarrow \ln \left(\frac{y}{P}\right)=\frac{rt}{100}$

$\Rightarrow \frac{y}{P}=e^{\frac{rt}{100}}$

$\therefore y=P{e}^{\frac{rt}{100}}$