$$L=\lim_{n\to \infty}\left(\;^nC_{0}\,\cdot\,^nC_{1} \,\cdot\,\cdots\,\cdot\,^nC_{n-1}\,\cdot\,^nC_{n}\;\right)^{1/(n(n+1))}$$
My attmept:
Taking log on both sides:
$$\ln L= \lim_{n\to \infty} \frac{\ln (^nC_{0}) +\ln(^nC_{1})+\cdots+\ln(^nC_{n-1})+\ln(^nC_{n})}{n(n+1)}$$
Applying L'hospital rule (differentiating numerator and denominator with respect to $n$) because it is $\dfrac{\infty}{\infty}$ form
$$\ln L=\lim_{n\to \infty}\frac{0+\frac{1}{n}+\frac{2!(2n-1)}{n(n-1)}+\cdots+\frac{1}{n}+0}{2n+1}=\frac{0}{\infty}=0 \implies L=1 $$
But answer is coming $L= \sqrt{e}$
i don't see where i'm doing wrong. in know how answer is coming $\sqrt{e}$ but
why L hospital rule isn't working .
i think there is no problem in differentiating function in numerator becuase all functions are continuos at $n =\infty$
Each term in your numerator $\to 0$, but the number of such terms is $n+1$, which $\to\infty$. So it's not obvious, for example, that their sum tends to $0$; indeed, it doesn't. If you want to solve the problem properly, I recommend the Stirling approximation $k!\approx\sqrt{2\pi}k^{k+1/2}e^{-k}$.