# How do I show that lim x->inf (1 + 1/n)^n = e ?

• May 6th 2010, 10:33 AM
s3a
How do I show that lim x->inf (1 + 1/n)^n = e ?
How do I SHOW that lim x->inf (1 + 1/x)^x = e ? Is it possible with techniques (easy and/or hard) (as long as it is using what schools teach Calculus 2 students - in other words, I'm bombarded with too much work at the moment to be able to learn something that is not part of my course)?

Any input would be greatly appreciated!
• May 6th 2010, 10:51 AM
Quote:

Originally Posted by s3a
How do I SHOW that lim x->inf (1 + 1/x)^x = e ? Is it possible with techniques (easy and/or hard) (as long as it is using what schools teach Calculus 2 students - in other words, I'm bombarded with too much work at the moment to be able to learn something that is not part of my course)?

Any input would be greatly appreciated!

You can use logs in conjunction with L'Hopital's rule

or you can use the binomial expansion.

$\displaystyle ln\left(1+\frac{1}{x}\right)^x=ln(limit)$

$\displaystyle xln\left(1+\frac{1}{x}\right)=ln(limit)$

$\displaystyle \frac{ln\left(1+\frac{1}{x}\right)}{\left(\frac{1} {x}\right)}=ln(limit)$

This gives 0/0 as x goes to infinity,
so applying L'Hopital's rule, differentiating numerator and denominator with the chain rule..

$\displaystyle \frac{\frac{1}{1+\frac{1}{x}}\left(-\frac{1}{x^2}\right)}{\left(-\frac{1}{x^2}\right)}$

$\displaystyle =\frac{1}{\frac{x+1}{x}}=\frac{x}{x+1}$

As x approaches infinity, this approaches 1,
hence

$\displaystyle 1=ln(limit)\ \Rightarrow\ e^1=limit$
• May 6th 2010, 11:02 AM
s3a
Just a quick question; is it acceptable to use the word "limit" as a variable on a test? What would you recommend?
• May 6th 2010, 11:29 AM
Quote:

Originally Posted by s3a
Just a quick question; is it acceptable to use the word "limit" as a variable on a test? What would you recommend?

Not as a variable, no...

I placed it in brackets for notational convenience as a shortcut,
however, it's preferable maybe to use a capital L.

$\displaystyle \lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right )^x=L$

then

$\displaystyle \lim_{x\rightarrow\infty}ln\left(1+\frac{1}{x}\rig ht)^x=ln(L)$

$\displaystyle \lim_{x\rightarrow\infty}xln\left(1+\frac{1}{x}\ri ght)=ln(L)$

$\displaystyle \lim_{x\rightarrow\infty}\frac{ln\left(1+\frac{1}{ x}\right)}{\left(\frac{1}{x}\right)}=ln(L)$

Then applying L'Hopital's rule, differentiating numerator and denominator..

$\displaystyle \lim_{x\rightarrow\infty}\frac{x+1}{x}=ln(L)$

$\displaystyle 1=ln(L)$

$\displaystyle L=e$

$\displaystyle \lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right )^x=e$
• May 6th 2010, 11:32 AM
Plato
Quote:

Originally Posted by s3a
How do I SHOW that lim x->inf (1 + 1/x)^x = e ? Is it possible with techniques (easy and/or hard) (as long as it is using what schools teach Calculus 2 students

I am not sure what all schools teach Caculus II students.
Here is the way I present the limit.
We know that $\displaystyle \frac{1}{1+n}<\ln\left(1+\frac{1}{n}\right)<\frac{ 1}{n}$.
From which we get $\displaystyle e^{\frac{n}{1+n}}<\left(1+\frac{1}{n}\right)^n<e$.
The limit is squeezed between $\displaystyle e~\&~e$.