# Thread: Limits that equal e

1. ## Limits that equal e

Ah, yes, this seems like the most famous mathematical term, I have heard before. I was looking at a function whose limit was e as it went to infinity, so I decided to read up about this sort of thing, and found out how amazing e really is, went on a few hour rant reading various things about it. never really clicked before . Now my question is, if you given a limit say $\displaystyle \lim_{n\to\infinty}f(n)$ and it contains some sort of ratio/fraction where n is in the denominator and involves something to the nth power, this is an observation I made, which, of course in math, can get you somewhere very wrong, the observation that if the function contains something along the lines of what I just suggested it will go to a limit equivalent to e?
as I was suggested this example earlier by redsox apart from the traditional "definition" of e

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

2. Think of it this way.

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

ln of both sides:

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

$\displaystyle ln(L)=1$

Therefore, $\displaystyle L=e$

3. According to the basic property of exponential...

$\displaystyle a^{b} = e^{b\cdot \ln a}$ (1)

... is...

$\displaystyle n^{\frac{1}{\ln n}} = e^{\frac{\ln n}{\ln n}} = e$ , $\displaystyle \forall n$ (2)

... so that it is not necessary that $\displaystyle n \rightarrow \infty$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by chisigma
According to the basic property of exponential...

$\displaystyle a^{b} = e^{b\cdot \ln a}$ (1)

... is...

$\displaystyle n^{\frac{1}{\ln n}} = e^{\frac{\ln n}{\ln n}} = e$ , $\displaystyle \forall n$ (2)

... so that it is not necessary that $\displaystyle n \rightarrow \infty$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
So this would work for this limit as well.

$\displaystyle \lim_{n\to\infty}n^{\frac{1}{\sqrt{n}}}$

by the property of the exponentional you stated

$\displaystyle n^{\frac{1}{\sqrt{n}}} = e^{\frac{ln\frac{1}{\sqrt{n}}}{ln\frac{1}{\sqrt{n} }}}$
would that still hold true? forgot the power of n, but you catch my drift

5. $\displaystyle n^{\frac{1}{\sqrt{n}}}\to1$. Informally $\displaystyle \sqrt{n}$ is much too "strong". It will completely overpower the base. I mean there are many more limits that give $\displaystyle e$. One that is not particularly interesting but comes to mind is $\displaystyle \lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=e$.

6. Originally Posted by Drexel28
$\displaystyle n^{\frac{1}{\sqrt{n}}}\to1$. Informally $\displaystyle \sqrt{n}$ is much too "strong". It will completely overpower the base. I mean there are many more limits that give $\displaystyle e$. One that is not particularly interesting but comes to mind is $\displaystyle \lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=e$.
Yes, would the function I just suggested go to infinity? after I had time to work it out I got $\displaystyle e^{\frac{ln(n)}{\sqrt{n}}}$