0
$\begingroup$

I have two questions:

$$\lim_{n\rightarrow \infty} \frac {(n!)^{\frac {1}{n}}}{n}$$ and $$\lim_{n\rightarrow \infty} \frac {1}{n}\ln {2n \choose n}$$

I realised that I had to do this with the help of integration, and in both cases I had to come to the same situation $\int_{0}^{1}\ln \space xdx$. But then I'm having $(x\times ln\space x-x)|_{0}^{1}$ and am stuck in the $0\times ln \space 0$ thing. What should I do with it?

$\endgroup$
4
0
$\begingroup$

A simple substitution lets you see whats happening. Substitute $x=1/y$ into the limit so that it becomes $$\frac{-ln(y)}{y}.$$ As x approaches zero y approaches infinity, and it's a little more intuitive that the above limit should go to zero. You can prove it with L'Hopital.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.