Limits that equal e

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November 20th 2009, 03:08 AM
RockHard

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 $\lim_{n\to\infinty}f(n)$ 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

$\lim_{n\to\infty}f(n)=n^{\frac{1}{ln(n)}}$
November 20th 2009, 03:48 AM
galactus

Think of it this way.

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

ln of both sides:

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

$ln(L)=1$

Therefore, $L=e$
November 20th 2009, 05:11 AM
chisigma

According to the basic property of exponential...

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

... is...

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

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

Kind regards

$\chi$ $\sigma$
November 20th 2009, 06:46 AM
RockHard

Quote:

Originally Posted by chisigma

According to the basic property of exponential...

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

... is...

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

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

Kind regards

$\chi$ $\sigma$

So this would work for this limit as well.

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

by the property of the exponentional you stated

$n^{\frac{1}{\sqrt{n}}} = e^{\frac{ln\frac{1}{\sqrt{n}}}{ln\frac{1}{\sqrt{n} }}}$
would that still hold true? forgot the power of n, but you catch my drift
November 20th 2009, 07:16 AM
Drexel28

$n^{\frac{1}{\sqrt{n}}}\to1$ . Informally $\sqrt{n}$ is much too "strong". It will completely overpower the base. I mean there are many more limits that give $e$ . One that is not particularly interesting but comes to mind is $\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=e$ .
November 20th 2009, 09:10 AM
RockHard

Quote:

Originally Posted by Drexel28

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

Yes, would the function I just suggested go to infinity? after I had time to work it out I got $e^{\frac{ln(n)}{\sqrt{n}}}$