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

**MechEng** · July 19th 2010, 01:42 PM

Good afternoon,

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

$ 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:

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

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

Where did this come from?

**Plato** · July 19th 2010, 02:16 PM

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

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

From that the limit is clear.

**adkinsjr** · July 19th 2010, 02:24 PM

Originally Posted by MechEng

Good afternoon,

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

$ 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:

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

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

Where did this come from?

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

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

Then take the limit of both sides.

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

Now use l-hospitals rule:

**CaptainBlack** · July 19th 2010, 02:37 PM

See this thread

CB

**Archie Meade** · July 19th 2010, 03:47 PM

Originally Posted by MechEng

Good afternoon,

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

$ 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:

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

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

Where did this come from?

The Binomial Expansion is

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

The Taylor Series expansion for $e^x$ is

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

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

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

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

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

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

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

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

**MechEng** · July 20th 2010, 09:13 AM

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

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?

**MechEng** · July 20th 2010, 10:30 AM

Originally Posted by adkinsjr

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

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

Then take the limit of both sides.

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

Now use l-hospitals rule:

Ah...

I think I follow this, however...

When we take the log of both sides and get:

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

Why are our new limits not:

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

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.

**Archie Meade** · July 20th 2010, 12:05 PM

Originally Posted by MechEng

Ah...

I think I follow this, however...

When we take the log of both sides and get:

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

Why are our new limits not:

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

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,

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

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.

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

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

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

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

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

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

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

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

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

$y=e^1$

Therefore..

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

**MechEng** · July 20th 2010, 12:37 PM

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...

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

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

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

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

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

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

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

**Archie Meade** · July 20th 2010, 01:26 PM

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...

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

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

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

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

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

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

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.

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

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

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

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

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