A limit
- Thread starter NRS
- Start date
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.
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.
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 \(\displaystyle \lim_{x\to 0}\left(1+\frac{1}{n}\right)^{n}=e\)
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
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- Mar 8, 2009
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\(\displaystyle One \ way.\)
\(\displaystyle a_n \ = \ \lim_{n\to\infty}\bigg(1+\frac{k}{n}\bigg)^n\)
\(\displaystyle Then \ ln|a_n| \ = \ ln\bigg[\lim_{n\to\infty}\bigg(1+\frac{k}{n}\bigg)^n\bigg], \ take \ natural \ logarithm \ of \ both \ sides\)
\(\displaystyle Ergo, \ ln|a_n| \ = \ \lim_{n\to\infty}ln\bigg(1+\frac{k}{n}\bigg)^n\bigg \ = \ \lim_{n\to\infty}nln\bigg(1+\frac{k}{n}\bigg), \ a_n \ is \ continuous \ as \ n\implies\infty\)
\(\displaystyle ln|a_n| \ = \ \lim_{n\to\infty}\frac{ln\bigg(1+\frac{k}{n}\bigg)}{\frac{1}{n}}, \ gives \ indeterminate \ form \ \frac{0}{0}, \ Marqui \ time.\)
\(\displaystyle ln|a_n| \ = \ \lim_{n\to\infty}\frac{kn}{n+k}, \ Marqui \ again, \ = \ \lim_{n\to\infty}\frac{k}{1} \ = \ k\)
\(\displaystyle Now, \ since \ ln|a_n| \ = \ k \ \implies \ a_n \ = \ e^k, \ QED\)
\(\displaystyle Note: \ I \ should \ have \ preface \ this, \ however \ I'll \ put \ it \ in \ here.\)
\(\displaystyle Limit \ of \ a \ Sequence: \ Let \ f \ be \ a \ function \ of \ a \ real \ variable \ such \ that \ \lim_{x\to\infty}f(x) \ = \ L\)
\(\displaystyle If \ \{a_n\} \ is \ a \ sequence \ such \ that \ f(n) \ = \ a_n \ for \ every \ positive \ integer \ n, \ then \ \lim_{n\to\infty}a_n \ = \ L\)
\(\displaystyle a_n \ = \ \lim_{n\to\infty}\bigg(1+\frac{k}{n}\bigg)^n\)
\(\displaystyle Then \ ln|a_n| \ = \ ln\bigg[\lim_{n\to\infty}\bigg(1+\frac{k}{n}\bigg)^n\bigg], \ take \ natural \ logarithm \ of \ both \ sides\)
\(\displaystyle Ergo, \ ln|a_n| \ = \ \lim_{n\to\infty}ln\bigg(1+\frac{k}{n}\bigg)^n\bigg \ = \ \lim_{n\to\infty}nln\bigg(1+\frac{k}{n}\bigg), \ a_n \ is \ continuous \ as \ n\implies\infty\)
\(\displaystyle ln|a_n| \ = \ \lim_{n\to\infty}\frac{ln\bigg(1+\frac{k}{n}\bigg)}{\frac{1}{n}}, \ gives \ indeterminate \ form \ \frac{0}{0}, \ Marqui \ time.\)
\(\displaystyle ln|a_n| \ = \ \lim_{n\to\infty}\frac{kn}{n+k}, \ Marqui \ again, \ = \ \lim_{n\to\infty}\frac{k}{1} \ = \ k\)
\(\displaystyle Now, \ since \ ln|a_n| \ = \ k \ \implies \ a_n \ = \ e^k, \ QED\)
\(\displaystyle Note: \ I \ should \ have \ preface \ this, \ however \ I'll \ put \ it \ in \ here.\)
\(\displaystyle Limit \ of \ a \ Sequence: \ Let \ f \ be \ a \ function \ of \ a \ real \ variable \ such \ that \ \lim_{x\to\infty}f(x) \ = \ L\)
\(\displaystyle If \ \{a_n\} \ is \ a \ sequence \ such \ that \ f(n) \ = \ a_n \ for \ every \ positive \ integer \ n, \ then \ \lim_{n\to\infty}a_n \ = \ L\)
BigGlenntheHeavy
Senior Member
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- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle What!!! \ You \ have \ never \ heard \ of \ "The \ Marqui \ Guillaume \ Francois \ Antoine \ de \ L'Hopital"?\)
\(\displaystyle My \ astonishment \ knows \ no \ bounds.\)
\(\displaystyle My \ astonishment \ knows \ no \ bounds.\)
