A limit

[NRS]

I can't seem to follow the process of taking this limit. I am testing for the convergence of a sequence:

[attachment=0:byto666p]DSC03623.jpg[/attachment:byto666p]

 

[galactus]

What is it you want to do?. Prove the limit exists?.

You can see that it does converge to e by plugging in larger and larger values of n.

But, to prove the limit exists takes a little more.

This is a very famous limit. Seen all across the math world in numerous places.

 

[NRS]

Oops, I probably should have been WAY more specific.

This is from the answer key, all I am given a[sub:184we4fh]n[/sub:184we4fh].

I just can't quite figure out how to solve the original equation as a limit.

 

[galactus]

We can get more rigorous, but try approaching it from the differentiability of ln(x) at x=1 using the definition of a derivative.

Since the derivative of ln(x) is 1 at x=1.

$\displaystyle \lim_{n\to 0}\frac{ln(1+n)-ln(1)}{n}=\lim_{n\to 0}\frac{ln(1+n)}{n}=\lim_{n\to 0} ln(1+n)^{\frac{1}{n}}=1$

Taking e to both sides gives:

$\displaystyle e=e^{\lim_{n\to 0} ln(1+n)^{\frac{1}{n}}}$

Since e is continuous, we have:

$\displaystyle e=\lim_{n\to 0} e^{ln(1+n)^{\frac{1}{n}}}=\lim_{n\to 0} (1+n)^{\frac{1}{n}}$

Now, just amke the sub they mentioned to get the other forms.


We couuld also use $\displaystyle \frac{1}{n+1}<\int_{n}^{n+1}\frac{1}{x}dx<\frac{1}{n}, \;\ n>0$

to show that $\displaystyle \lim_{x\to 0}\left(1+\frac{1}{n}\right)^{n}=e$

 

[BigGlenntheHeavy]

$\displaystyle One \ way.$

$\displaystyle a_n \ = \ \lim_{n\to\infty}\bigg(1+\frac{k}{n}\bigg)^n$

$\displaystyle Then \ ln|a_n| \ = \ ln\bigg[\lim_{n\to\infty}\bigg(1+\frac{k}{n}\bigg)^n\bigg], \ take \ natural \ logarithm \ of \ both \ sides$

\(\displaystyle Ergo, \ ln|a_n| \ = \ \lim_{n\to\infty}ln\bigg(1+\frac{k}{n}\bigg)^n\bigg \ = \ \lim_{n\to\infty}nln\bigg(1+\frac{k}{n}\bigg), \ a_n \ is \ continuous \ as \ n\implies\infty\)

$\displaystyle ln|a_n| \ = \ \lim_{n\to\infty}\frac{ln\bigg(1+\frac{k}{n}\bigg)}{\frac{1}{n}}, \ gives \ indeterminate \ form \ \frac{0}{0}, \ Marqui \ time.$

$\displaystyle ln|a_n| \ = \ \lim_{n\to\infty}\frac{kn}{n+k}, \ Marqui \ again, \ = \ \lim_{n\to\infty}\frac{k}{1} \ = \ k$

$\displaystyle Now, \ since \ ln|a_n| \ = \ k \ \implies \ a_n \ = \ e^k, \ QED$

$\displaystyle Note: \ I \ should \ have \ preface \ this, \ however \ I'll \ put \ it \ in \ here.$

$\displaystyle Limit \ of \ a \ Sequence: \ Let \ f \ be \ a \ function \ of \ a \ real \ variable \ such \ that \ \lim_{x\to\infty}f(x) \ = \ L$

$\displaystyle If \ \{a_n\} \ is \ a \ sequence \ such \ that \ f(n) \ = \ a_n \ for \ every \ positive \ integer \ n, \ then \ \lim_{n\to\infty}a_n \ = \ L$

 

[NRS]

Thankyou, but what do you mean when you say "Marqui"?

BTW What do you use to get the graphical functions up there?

 

[BigGlenntheHeavy]

$\displaystyle What!!! \ You \ have \ never \ heard \ of \ "The \ Marqui \ Guillaume \ Francois \ Antoine \ de \ L'Hopital"?$

$\displaystyle My \ astonishment \ knows \ no \ bounds.$

 

[NRS]

I guess I'm just just not familiar enough with him to be on a first name basis yet...

Thanks!

P.S. I thought that looked misteriously like L^Hopital!