Thread: lim (1+1/x)^x as x->infinity

Good afternoon,

I am working on a problem that asks to find the following limit:

$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$

This should be equal to e, but I am having some difficulty remembering why.

I checked WolframAlpha and was provided with the following:

$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$

Indeterminate form of type $\displaystyle 1^\infty$. Transform using
$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$ = $\displaystyle
lim_{e^{x\rightarrow\infty}}x log(1+\frac{1}{x})
$


Where did this come from?

Using Napier’s inequality it is easy to show that:
$\displaystyle \displaystyle \frac{1}{n+1}\le\log \left( {1 + \frac{1}{n}} \right) \leqslant \frac{1}{n} $.

So $\displaystyle e^{\frac{1}{{n + 1}}} \leqslant \left( {1 + \frac{1}{n}} \right) \leqslant e^{\frac{1}{n}} \; \Rightarrow \;e^{\frac{n}{{n + 1}}} \leqslant \left( {1 + \frac{1}{n}} \right)^n \leqslant e$

From that the limit is clear.

Originally Posted by MechEng

Good afternoon,

I am working on a problem that asks to find the following limit:

$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$

This should be equal to e, but I am having some difficulty remembering why.

I checked WolframAlpha and was provided with the following:

$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$

Indeterminate form of type $\displaystyle 1^\infty$. Transform using
$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$ = $\displaystyle
lim_{e^{x\rightarrow\infty}}x log(1+\frac{1}{x})
$


Where did this come from?

If you let $\displaystyle y=\left(1+\frac{1}{x}\right)^x$ Then take the log of both sides you can write:

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

Then take the limit of both sides.

$\displaystyle ln(\lim_{x->\infty}y)=\lim_{x->\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac {1}{x}}$

Now use l-hospitals rule:

http://www.youtube.com/watch?v=6g_g9-AU188

See this thread

CB

Originally Posted by MechEng

Good afternoon,

I am working on a problem that asks to find the following limit:

$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$

This should be equal to e, but I am having some difficulty remembering why.

I checked WolframAlpha and was provided with the following:

$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$

Indeterminate form of type $\displaystyle 1^\infty$. Transform using
$\displaystyle
lim_{x\rightarrow\infty}(1+\frac{1}{x})^x
$ = $\displaystyle
lim_{e^{x\rightarrow\infty}}x log(1+\frac{1}{x})
$


Where did this come from?

The Binomial Expansion is

$\displaystyle (a+b)^n=\binom{n}{0}a^nb^0+\binom{n}{1}a^{n-1}b^1+\binom{n}{2}a^{n-2}b^2+....+\binom{n}{n}a^{n-n}b^n$

The Taylor Series expansion for $\displaystyle e^x$ is

$\displaystyle e^x=\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}=1+x+\frac{x^2}{2 !}+\frac{x^3}{3!}+.....$

Using the Binomial Expansion for $\displaystyle \left(1+\frac{1}{x}\right)^x$

$\displaystyle \left(1+\frac{1}{x}\right)^x=\binom{x}{0}+\binom{x }{1}\frac{1}{x}+\binom{x}{2}\frac{1}{x^2}+\binom{x }{3}\frac{1}{x^3}+......$

$\displaystyle =1+1+\frac{x(x-1)}{2!}\frac{1}{x^2}+\frac{x(x-1)(x-2)}{3!}\frac{1}{x^3}+....$

$\displaystyle =1+1+\frac{x^2-x}{2!x^2}+\frac{x^3-3x^2+2x}{3!x^3}+.....$

Using the fact that $\displaystyle \displaystyle \lim_{x\ \rightarrow\ \infty}\frac{1}{x^n}=0,\ n=1,\ 2,\ 3,\ 4,\ ......$

and all the powers of x will be in the denominator position above,
therefore as x goes to infinity, it reduces to

$\displaystyle \displaystyle \lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right )^x=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\fr ac{1}{5!}+.....=e^1$

I am still somewhat confused by this. I can see that e is defined as $\displaystyle e=lim_{n\rightarrow\infty}(1+\frac{1}{n})^n$.

I think I might be having another mental lapse with respect finding this limit. I am working on sequences, so is there some corelation there that I am missing?

Originally Posted by adkinsjr

If you let $\displaystyle y=\left(1+\frac{1}{x}\right)^x$ Then take the log of both sides you can write:

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

Then take the limit of both sides.

$\displaystyle ln(\lim_{x->\infty}y)=\lim_{x->\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac {1}{x}}$

Now use l-hospitals rule:

YouTube - L'Hospital's Rule - Indeterminate Powers

Ah...

I think I follow this, however...

When we take the log of both sides and get:

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

Why are our new limits not:

$\displaystyle ln(\lim_{x->\infty}y)=xln\lim_{x->\infty}(1+\frac{1}{x}\right)$

Is the ln treated as a constant?

I hope my question is not too rediculous.

PS... I would love to watch your youtube video, but youtube is blocked at my place of work.

Originally Posted by MechEng

Ah...

I think I follow this, however...

When we take the log of both sides and get:

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

Why are our new limits not:

$\displaystyle ln(\lim_{x->\infty}y)=xln\lim_{x->\infty}(1+\frac{1}{x}\right)$

Is the ln treated as a constant?

I hope my question is not too rediculous.

PS... I would love to watch your youtube video, but youtube is blocked at my place of work.

Firstly,

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

If we take the natural logarithms of both sides, we will be able to handle the "index" x before evaluating the limit using L'Hopital's rule.

$\displaystyle log_ey=log_e\left(1+\frac{1}{x}\right)^x=xlog_e\le ft(1+\frac{1}{x}\right)=\frac{1}{\left(\frac{1}{x} \right)}log_e\left(1+\frac{1}{x}\right)=\frac{log_ e\left(1+\frac{1}{x}\right)}{\left(\frac{1}{x}\rig ht)}$

This brings y into a form that can be evaluated as $\displaystyle x\rightarrow\ \infty$ using L"Hopital's Rule, since if $\displaystyle x\ \rightarrow\ \infty$

$\displaystyle \frac{log_e\left(1+\frac{1}{x}\right)}{\left(\frac {1}{x}\right)}$ evaluates to

$\displaystyle \frac{log_e1}{0}\ \rightarrow\ \frac{0}{0}$

Hence, differentiating numerator and denominator to apply L'Hopital's Rule..

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

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

Now this can be evaluated as x approaches infinity....

$\displaystyle log_ey=\frac{1}{1+0}$ as $\displaystyle x\ \rightarrow\ \infty$

$\displaystyle y=e^1$

Therefore..

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

First off, thank you all for your continued efforts here.

Secondly, I apologize for being so slow to catch on here, requiring everyone to reiterate.

However, I think I have a more clear and concise understanding of where I'm losing you...

$\displaystyle log_ey=log_e\left(1+\frac{1}{x}\right)^x$


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


Right about here... how (or why, for that matter) did we end up with $\displaystyle x=\frac{1}{(\frac{1}{x})}$ ?

Are we simply jostling the terms about to allow us to apply L'Hopital's rule?


$\displaystyle =\frac{1}{\left(\frac{1}{x}\right)}log_e\left(1+\f rac{1}{x}\right)$


$\displaystyle =\frac{log_e\left(1+\frac{1}{x}\right)}{\left(\fra c{1}{x}\right)}$

I appologize if this process has been frustrating... it has been very taxing for me.

Originally Posted by MechEng

First off, thank you all for your continued efforts here.

Secondly, I apologize for being so slow to catch on here, requiring everyone to reiterate.

However, I think I have a more clear and concise understanding of where I'm losing you...

$\displaystyle log_ey=log_e\left(1+\frac{1}{x}\right)^x$


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


Right about here... how (or why, for that matter) did we end up with $\displaystyle x=\frac{1}{(\frac{1}{x})}$ ?

Are we simply jostling the terms about to allow us to apply L'Hopital's rule?


$\displaystyle =\frac{1}{\left(\frac{1}{x}\right)}log_e\left(1+\f rac{1}{x}\right)$


$\displaystyle =\frac{log_e\left(1+\frac{1}{x}\right)}{\left(\fra c{1}{x}\right)}$

I appologize if this process has been frustrating... it has been very taxing for me.

Yes, you could say we are jostling terms here in order to reach a suitable form to apply the rule.

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

hence, $\displaystyle x=\frac{x\left(\frac{1}{x}\right)}{\left(\frac{1}{ x}\right)}=\frac{\left(\frac{x}{x}\right)}{\left(\ frac{1}{x}\right)}$

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

Take your time to understand.
There are numerous alternative ways to tackle the problem,
including that very interesting inequality pointed out by Plato.