Proof What is the limit of(1+(1/x))^x as x approaches infinity?

[ksigel]

ln y= lim (x->00) ln (1+(1/x))/x^-1 = f(x) = ln x+1/x
f'(x)=x/(x+1)= 1/(1+(1/x))
g'(x)= -x^-2
ln y= lim (x->00) 1/(1+(1/x))/x^-2 (L'Hopital's Rule)
However, the published proof has an intermediate step ln y= lim (x->00) 1/(1+(1/x)) (0-x^-2)/-x^-2
Where does the term (0-x^-2) come from?
Thanks
For complete proof see https://socratic.org/questions/what-is-the-limit-of-1-1-x-x-as-x-approaches-infinity-2

 

[Deleted member 4993]

ln y= lim (x->00) ln (1+(1/x))/x^-1 = f(x) = ln x+1/x
f'(x)=x/(x+1)= 1/(1+(1/x))
g'(x)= -x^-2
ln y= lim (x->00) 1/(1+(1/x))/x^-2 (L'Hopital's Rule)
However, the published proof has an intermediate step ln y= lim (x->00) 1/(1+(1/x)) (0-x^-2)/-x^-2
Where does the term (0-x^-2) come from?
Thanks
For complete proof see https://socratic.org/questions/what-is-the-limit-of-1-1-x-x-as-x-approaches-infinity-2

d/dx[1+1/x] = [0 - 1/x²]

 

[lex]

$\dfrac{d}{dx} \ln h(x) = \dfrac{h'(x)}{h(x)}=\dfrac{1}{h(x)}\times h'(x) \hspace3ex$ (*)

If $h(x) = 1+\tfrac{1}{x}=1+x^{-1}$
then $\hspace5ex h'(x)=0-x^{-2}$

$\therefore \dfrac{d}{dx} \ln \left(1+\tfrac{1}{x}\right)=\dfrac{1}{\left(1+\tfrac{1}{x}\right)}\times (0-x^{-2}) \hspace3ex$ by (*)

 

[ksigel]

Many thanks!

 

[ksigel]

Thank you for your help.
I have an additional question: the proof of the limit of (1+(1/x))^x as x approaches infinity uses the natural logarithm to find the solution. However, it seems to me that if a logarithm of a different base is used, the limit would approach that base. Is the proof dependent upon the base chosen?

 

[lex]

This question was about finding a limit:
$\lim \limits_{x \to \infty} \left(1+\tfrac{1}{x}\right)^x$
What limit are you now looking to find?

 

[ksigel]

The same limit. The proof involves taking the natural logarithm of both sides. But why the natural logarithm? Why not the logarithm with any other base?

 

[Deleted member 4993]

The same limit. The proof involves taking the natural logarithm of both sides. But why the natural logarithm? Why not the logarithm with any other base?

Now that you know the process - you show us what would happen if take "log" with some other base.

 

[lex]

The same limit. The proof involves taking the natural logarithm of both sides. But why the natural logarithm? Why not the logarithm with any other base?

So you know that $\lim \limits_{x \to \infty} \left(1+\tfrac{1}{x}\right)^x = e$
It doesn't matter how you prove it, you still obviously get $e$.

So your question is (i) why did we use natural log rather than e.g. log to the base 10, when proving that $\lim \limits_{x \to \infty} \left(1+\tfrac{1}{x}\right)^x = e$
and possibly (ii) what would happen in the proof if we used a different base for the log?

The answer to (i) is that the proof involves differentiating log x, which is easy to do for natural log:
$\tfrac{d}{dx} \ln f(x) = \tfrac{f'(x)}{f(x)}$
However $\tfrac{d}{dx} \log_a f(x) = (\log_a e) \tfrac{f'(x)}{f(x)} \hspace2ex$, a>0
As @Subhotosh Khan suggests, you should be able to complete the proof using a different base, and show that the answer is '$e$' no matter what base we use in the proof.

 

[ksigel]

Attached please find my "proof" using base 10. Where is my error?

 

[ksigel]

Many thanks for your explanation Lex. All roads lead to Rome (or in this case, all bases lead to e). Can you prove this for me please?

 

[pka]

The same limit. The proof involves taking the natural logarithm of both sides. But why the natural logarithm? Why not the logarithm with any other base?

It may surprise you but the following is true:$\Large\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{{bx + c}}} \right)^{dx}} = {e^{\frac{{ad}}{b}}}.$

 

[ksigel]

Thanks for the response pka. But how does this fact help answer my question (why all bases lead to e)?
BTW- how do you generate those symbols? When I copy and paste, I get: lim⁡x→∞(1+abx+c)dx=eadb.\Large\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{{bx + c}}} \right)^{dx}} = {e^{\frac{{ad}}{b}}}.x→∞lim(1+bx+ca)dx=ebad.

 

[lookagain]

However $\tfrac{d}{dx} \log_a f(x) = (\log_a e) \tfrac{f'(x)}{f(x)} \hspace2ex$, a>0

And $\displaystyle \ a \ne 1$.

 

[lex]

And $\displaystyle \ a \ne 1$.

Yes - thanks, of course.

 

[lex]

First note, applying the logarithm change of base rule:
$\ln x = \dfrac{\log_a x}{\log_a e} \hspace3ex$ (1)

$y=\lim \limits_{x \to \infty}\left(1+\tfrac{1}{x}\right)^x \hspace3ex$ assuming the limit exists (a*)

$\log_a y =\log_a \lim \limits_{x \to \infty}\left(1+\tfrac{1}{x}\right)^x \hspace3ex$ assuming limit >0, 1≠a>0 (b*)

$\log_a y = \lim \limits_{x \to \infty} \log_a \left(1+\tfrac{1}{x}\right)^x \hspace3ex$ this requires justification (omitted in your linked proof) (c*)

$\log_a y = \lim \limits_{x \to \infty} (\log_a e) \ln\left(1+\tfrac{1}{x}\right)^x \hspace3ex$ by (1)

$\log_a y = (\log_a e) \lim\limits_{x \to \infty} \left(1+\tfrac{1}{x}\right)^x$

Now everything follows in your original proof, with the RHS $\times$ the constant $(\log_a e)$
... until
$\log_a y = (\log_a e).1\\ \rightarrow y=a^{\log_a e} = e$

To deal with (a*), (b*), (c*), first observe:

$\left(1+\tfrac{1}{x}\right)^x >1$ for all x>0 (2)

for a(*) it can be shown fairly easily that $\left(1+\tfrac{1}{x}\right)^x$ is an increasing function on $(1,\infty)$ and bounded above by $e$ (point (2) above), therefore the limit exists

for (b*) (point (2) above) $\left(1+\tfrac{1}{x}\right)^x >1$ for all x>0, therefore if the limit exists, $\lim\limits_{x \to \infty} \left(1+\tfrac{1}{x}\right)^x=k≥1>0$

for (c*)
$\log_a x$ is continuous at $x=k>0$
so $\log_a \lim\limits_{u \to k} u = \lim\limits_{u \to k} \log_a u$
$\therefore \log_a \lim \limits_{x \to \infty}\left(1+\tfrac{1}{x}\right)^x = \lim \limits_{x \to \infty} \log_a \left(1+\tfrac{1}{x}\right)^x$



Btw in your work, you assumed incorrectly that $\tfrac{d}{dx} \log_a f(x) = \tfrac{f'(x)}{f(x)} \hspace2ex e,1≠a>0$
As I pointed out:
$\tfrac{d}{dx} \log_a f(x) = (\log_a e) \tfrac{f'(x)}{f(x)} \hspace2ex 1≠a>0$

Also, to get the symbolic form to appear you must open and close 'imath' tags. Note, they need to be lower-case letters. (I am making them upper-case, so they are not interpreted)
[IMATH]\Large\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{{bx + c}}} \right)^{dx}} = {e^{\frac{{ad}}{b}}}[/IMATH]

 

[ksigel]

Lex,
Thank you for your extensive reply. I appreciate the time you took to help me. It will take me some time for me to assimilate your answer. Your answer has generated two new questions for me:
1) Can you provide a proof for the fourth equation please (the equation that requires justification)?
2) How do I open imath tags?
Many thanks.

 

[pka]

2) How do I open imath tags?

ksigel, do you know how to code in LaTeX ? if you do then on the menu bar there is $f(x)$ icon.
Click on it when the input window appears select the inline option and inter your code. When you are satisfied click continue.

 

[ksigel]

ksigel, do you know how to code in LaTeX ? if you do then on the menu bar there is $f(x)$ icon.
Click on it when the input window appears select the inline option and inter your code. When you are satisfied click continue.

Don't know a think about LaTeX. Is it hard to learn? How do I begin?

 

[lex]

This was my proof for (c*):

$\log_a x$ is continuous at $x=k>0$
so $\log_a \lim\limits_{u \to k} u = \lim\limits_{u \to k} \log_a u$
$\therefore \log_a \lim \limits_{x \to \infty}\left(1+\tfrac{1}{x}\right)^x = \lim \limits_{x \to \infty} \log_a \left(1+\tfrac{1}{x}\right)^x$

Expanding on this.
To be able to 'reverse the order of the function and the limit', we want the limit to exist and the function to be continuous at the value of the limit.
You can use the ready-made theorem:

In the present case, $f(X)=\log X$ and $g(x)=\left(1+\tfrac{1}{x}\right)^x$
Unfortunately $\log X$ is not continuous at X=0, so we have to show that $\lim \limits_{x \to \infty}\left(1+\tfrac{1}{x}\right)^x$ exists and $=k>0$.

$\left(1+\tfrac{1}{x}\right)^x >1$ for all x>0, therefore if the limit exists, $\lim\limits_{x \to \infty} \left(1+\tfrac{1}{x}\right)^x=k≥1>0$
The only remaining problem is to prove that the limit exists.

It is sufficient to prove that $\left(1+\tfrac{1}{x}\right)^x$ is increasing, and bounded above by $e$ as there is a theorem which says that an increasing function, bounded above, converges to a limit.

$X>1$

$\ln X+\dfrac{1}{X}-1$ is strictly increasing and therefore $>0$, when $X>1\hspace3ex (\text{the derivative }\tfrac{1}{X}-\tfrac{1}{X^2}>0, X>1 \text{ and }\ln 1+\tfrac{1}{1}-1=0$)

$\therefore \ln(\tfrac{x+1}{x})+\tfrac{x}{x+1}-1 >0$ (decreasing), x>0 (since $X=\tfrac{x+1}{x} >1$, (decreasing), $x>0$)

i.e. $\ln(1+\tfrac{1}{x})-\tfrac{1}{1+x}>0$, (decreasing), $x>0$

i.e. $\dfrac{\text{d}}{\text{dx}}[x\ln(1+\tfrac{1}{x})]>0$, (decreasing), $x>0$

$\therefore \ln (1+\tfrac{1}{x})^x$ increasing, $x>0$

$\therefore (1+\tfrac{1}{x})^x$ increasing, $x>0\hspace3ex$ (since $e^x$ increasing)



Now $e^x-x-1$ is strictly increasing, $x>0\hspace2ex (\text{the derivative }e^x-1>0, \;x>0)$

$\rightarrow e^x-x-1>0, \;x>0 \hspace2ex (\text{since strictly increasing and } e^0-0-1=0)$

$e^{1/x}-\tfrac{1}{x}-1>0, \;x>0\\ \therefore e^{1/x}>1+\tfrac{1}{x}, \;x>0$

$(1+\tfrac{1}{x})^x<e,\; x>0$

As for Latex, you're probably best looking at examples that others have done. You should be able to right-click on a formula and show the Latex. However due to a recent upgrade on this site, this doesn't work at the moment. You can use it on other sites though to see their Latex.
You can see the Latex on this site by hitting the reply button however.