# Thread: limit of xth root etc etc

1. ## limit of xth root etc etc

Determine if $\displaystyle \lim_{x\to\infty} \frac{\sqrt[x]{x!}}{x}$ exist (and value) and if it doesn't explain why.
(in case thats too small, it reads the xth root of x! all over x)

Potentially VERY easy series questions?!?

Thanks

2. Originally Posted by polymerase
Determine if $\displaystyle \lim_{x\to\infty} \frac{\sqrt[x]{x!}}{x}$ exist (and value) and if it doesn't explain why.
(in case thats too small, it reads the xth root of x! all over x)

Potentially VERY easy series questions?!?

Thanks
use the fact taht $\displaystyle n!\sim{\sqrt{2\pi{n}}n^{n}e^{-n}}$

then it should be apparent

3. Originally Posted by Mathstud28
use the fact taht $\displaystyle n!\sim{\sqrt{2\pi{n}}n^{n}e^{-n}}$

then it should be apparent
Firstly, thank you

Secondly, so is the answer $\displaystyle \frac{1}{e}$?

Thirdly, is there another way to do this, because we never learned the relationship between n! and the stuff you stated above

4. Originally Posted by polymerase
Firstly, thank you

Secondly, so is the answer $\displaystyle \frac{1}{e}$?

Thirdly, is there another way to do this, because we never learned the relationship between n! and the stuff you stated above
Good job! Indeed the answer is $\displaystyle \frac{1}{e}$

and you could also do this $\displaystyle y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}$

and this is painful but $\displaystyle \ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}$

Or you could als use that $\displaystyle n!=\int_0^{\infty}x^{n}e^{-x}dx$

The gamma function and go L'hopitals

5. Originally Posted by Mathstud28
Good job! Indeed the answer is $\displaystyle \frac{1}{e}$

and you could also do this $\displaystyle y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}$

and this is painful but $\displaystyle \ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}$

Or you could als use that $\displaystyle n!=\int_0^{\infty}x^{n}e^{-x}dx$

The gamma function and go L'hopitals
REALLY?! thats the alternative! how the %$!@ does our prof. expect us to solve that? we never learn ANY of that! LOL is there ANOTHER way? 6. Originally Posted by Mathstud28 Good job! Indeed the answer is$\displaystyle \frac{1}{e}$and you could also do this$\displaystyle y=\lim_{x\to\infty}\frac{(x!)^{\frac{1}{x}}}{x}\Ri ghtarrow{\ln(y)=\frac{\ln(x!)}{x^2}}$and this is painful but$\displaystyle \ln(x!)\approx{x\ln(x)-x+\frac{\ln(x)}{2}+\frac{\ln(2\pi)}{2}}$Or you could als use that$\displaystyle n!=\int_0^{\infty}x^{n}e^{-x}dx$The gamma function and go L'hopitals Expanding upon my first thought after simplifying you would have$\displaystyle \lim_{n\to\infty}\frac{\ln(x)}{x}-1+\frac{\ln(x)}{2x^2}+\frac{\ln(2\pi)}{2x^2}=-1$therefore$\displaystyle \ln(y)=-1\Rightarrow{y=e^{-1}}$This came from the ln(x!) approximation over x² 7. Originally Posted by polymerase REALLY?! thats the alternative! how the %$!@ does our prof. expect us to solve that? we never learn ANY of that!

LOL is there ANOTHER way?
Haha...Uhm..I am surprised I thought of three ways...uhm...I cant think of anymore...maybe if you look at this post above you cna use that...that is not overly complicated? haha

8. Originally Posted by Mathstud28
Expanding upon my first thought after simplifying you would have $\displaystyle \lim_{n\to\infty}\frac{\ln(x)}{x}-\frac{1}{x}+\frac{\ln(x)}{2x^2}+\frac{\ln(2\pi)}{2 x^2}=-1$ therefore $\displaystyle \ln(y)=-1\Rightarrow{y=e^{-1}}$

This came from the ln(x!) approximation over x²
How do you figure out/calculate that $\displaystyle \ln (x!)=$above stuff?

9. Originally Posted by polymerase
How do you figure out/calculate that $\displaystyle \ln (x!)=$above stuff?
Uhm....well I have the memorized but if you need to see the entire thing...hold on let me find...an ariticle...ah! here we go read here

Natural logarithm - Wikipedia, the free encyclopedia

sorry edit:
Look here
Factorial - Wikipedia, the free encyclopedia

10. Originally Posted by Mathstud28
Or you could als use that $\displaystyle n!=\int_0^{\infty}x^{n}e^{-x}dx$
How will you use this?

11. Originally Posted by polymerase

LOL is there ANOTHER way?
\displaystyle \begin{aligned} \frac{\sqrt[x]{x!}}{x}&=\exp \left[ \ln \left( \prod\limits_{k\,=\,1}^{x}{\left( \frac{k}{x} \right)^{1/x}} \right) \right] \\ & =\exp \left[ \frac{1}{x}\sum\limits_{k\,=\,1}^{x}{\ln \frac{k}{x}} \right] \\ & =\exp \left[ \int_{0}^{1}{\ln y\,dy} \right],\,\text{as }x\to\infty. \end{aligned}

12. Nice limit question. Using my intuition I thought it would tend to positive infinite.