# infinite Limit

• Aug 22nd 2010, 08:26 AM
Tclack
infinite Limit
So I'm supposed to prove that
$\displaystyle lim_n_\rightharpoonup_\infty$$\displaystyle \frac{\sqrt[n]{n!}}{n} And I'm doing so by using the squeezing theorem (with guided help from the book) I know for certain that \displaystyle \frac{1}{e^(1-\frac{1}{n})} < \frac{\sqrt[n]{n!}}{n} < \frac{(1+\frac{1}{n})(n+1)^\frac{1}{n}}{e} (btw the left is 1 over e to the power of all that stuff, Latex is making it look weird) This is correct. My book confirmed so. And I can see how the left side of the equation approaches 1/e. Somehow the right side does and I can't see how....anybody? • Aug 22nd 2010, 08:46 AM Jester Quote: Originally Posted by Tclack So I'm supposed to prove that \displaystyle lim_n_\rightharpoonup_\infty$$\displaystyle \frac{\sqrt[n]{n!}}{n}$

And I'm doing so by using the squeezing theorem (with guided help from the book)

I know for certain that

$\displaystyle \frac{1}{e^(1-\frac{1}{n})} < \frac{\sqrt[n]{n!}}{n} < \frac{(1+\frac{1}{n})(n+1)^\frac{1}{n}}{e}$

(btw the left is 1 over e to the power of all that stuff, Latex is making it look weird)

This is correct. My book confirmed so. And I can see how the left side of the equation approaches 1/e. Somehow the right side does and I can't see how....anybody?

If you let $\displaystyle x = \frac{1}{n}$ then $\displaystyle \displaystyle \lim_{x \to 0} \frac{(1+x)(1+\frac{1}{x})^x}{e} = \frac{1}{e} \lim_{x \to 0} (1+x) \cdot \lim_{x \to 0} (1+\frac{1}{x})^x$.
• Aug 22nd 2010, 08:48 AM
Unknown008
For the LaTeX code, you should wrap the whole 'power' in brackets like this:

e^{(1 - \frac{1}{n})}

As for the right side, 1/infinity tends to 0, and anything to the power of zero becomes 1.

So, you are left with:

$\displaystyle \frac{1+\frac{1}{n}}{e}$

Once again, 1/infinity tends to zero, and 1 + 0 = 1. So, you get 1/e.

I hope it helps! (Happy)
• Aug 22nd 2010, 08:55 AM
Prove It
$\displaystyle \frac{\left(1 + \frac{1}{n}\right)(1 + n)^{\frac{1}{n}}}{e} = \frac{1}{e}\cdot e^{\ln{\left[\left(1 + \frac{1}{n}\right)(1 + n)^{\frac{1}{n}}\right]}}$

$\displaystyle = \frac{1}{e}\cdot e^{\ln{\left(1 + \frac{1}{n}\right)} + \frac{1}{n}\cdot\ln{(1 + n)}}$

$\displaystyle = \frac{1}{e}\cdot e^{\ln{\left(1 + \frac{1}{n}\right)}}\cdot e^{ \frac{\ln{(1 + n)}}{n}}$.

Therefore

$\displaystyle \lim_{n \to \infty}\frac{\left(1 + \frac{1}{n}\right)(1 + n)^{\frac{1}{n}}}{e} = \lim_{n \to \infty}\frac{1}{e}\cdot e^{\ln{\left(1 + \frac{1}{n}\right)}}\cdot e^{\frac{\ln{(1 + n)}}{n}}$

$\displaystyle = \frac{1}{e}\lim_{n \to \infty}e^{\ln{\left(1 + \frac{1}{n}\right)}}\lim_{n \to \infty}e^{\frac{\ln{(1 + n)}}{n}}$

$\displaystyle = \frac{1}{e}\cdot e^{\ln{1}} \cdot e^{\lim_{n \to \infty}\frac{\ln{(1+n)}}{n}}$

$\displaystyle = \frac{1}{e} \cdot e^{\lim_{n \to \infty}\frac{\frac{1}{1+n}}{1}}$ by L'Hospital's Rule

$\displaystyle = \frac{1}{e} \cdot e^{\lim_{n \to \infty}\frac{1}{1+n}}$

$\displaystyle = \frac{1}{e} \cdot e^0$

$\displaystyle = \frac{1}{e}$.
• Aug 22nd 2010, 09:15 AM
Tclack
Man, you guys replied fast, I was playing around with this for about 4 hours.

I'm trying to get better. Does anyone know of a book or source of really hard problems? I've found this:

When you have indeterminate limits (like in this case, $\displaystyle \infty^{0}$) you need to apply a transformation to get $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ so that you can then apply L'Hospital's Rule.