# limit proof

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• December 2nd 2010, 01:41 PM
pflo
limit proof
One way determine the deriveative of $e^x$ is to use inverse functions (natural logs). Another way is to use the definition of a derivative. Thus, $\frac{d}{dx}e^x=\lim_{h\rightarrow0}\frac{e^{x+h}-e^x}{h}$. When simplified, this results in the equation $\frac{d}{dx}e^x=e^x \lim_{h\rightarrow0}\frac{e^h-1}{h}$.

My questions is: what is a proof that $\lim_{h\rightarrow0}\frac{e^h-1}{h}=1$?
• December 2nd 2010, 02:00 PM
Quote:

Originally Posted by pflo
One way to determine the derivative of $e^x$ is to use inverse functions (natural logs). Another way is to use the definition of a derivative. Thus, $\frac{d}{dx}e^x=\lim_{h\rightarrow0}\frac{e^{x+h}-e^x}{h}$. When simplified, this results in the equation $\frac{d}{dx}e^x=e^x \lim_{h\rightarrow0}\frac{e^h-1}{h}$.

My question is: what is a proof that $\lim_{h\rightarrow0}\frac{e^h-1}{h}=1$?

$\displaystyle\ e^h=1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+.......$

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

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

then take the limit as $h\rightarrow\ 0$
• December 2nd 2010, 05:37 PM
alexandros
Quote:

$\displaystyle\ e^h=1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+.......$

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

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

then take the limit as $h\rightarrow\ 0$

To get the above you use the Taylor's expansion where :

f'(0) = $\frac{d(e^x)}{dx}$ at x=0 ,which is 1

f"(0) = $\frac{d^2(e^x)}{dx^2}$ at x=0 ,which is again 1 e.t.c .....e.t.c

Thus admitting that $\frac{d(e^x)}{dx}=e^x$

But that is what you want to prove
• December 3rd 2010, 01:18 AM
Quote:

Originally Posted by alexandros
To get the above you use the Taylor's expansion where :

f'(0) = $\frac{d(e^x)}{dx}$ at x=0 ,which is 1

f"(0) = $\frac{d^2(e^x)}{dx^2}$ at x=0 ,which is again 1 e.t.c .....e.t.c

Thus admitting that $\frac{d(e^x)}{dx}=e^x$

But that is what you want to prove

If the Taylor expansion is used, it can be considered "circular reasoning"
from one perspective.

Instead if we use the Binomial Expansion (Newton's Generalised Binomial Theorem) on

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

we obtain

$e=\binom{x}{0}+\binom{x}{1}\frac{1}{x}+\binom{x}{2 }\frac{1}{x^2}+.....$

from which we can write the power series expansion of $e^h$
• December 3rd 2010, 01:41 AM
alexandros
Is h a real or a natural No??
• December 3rd 2010, 01:58 AM
Quote:

Originally Posted by alexandros
Is h a real or a natural No??

$h\rightarrow\ 0$

$\displaystyle\ e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\rig ht)^n$

$\displaystyle\ e^h=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\r ight)^{nh}$

$\displaystyle\ =\lim_{n\rightarrow\infty}\left[\binom{nh}{0}+\binom{nh}{1}\frac{1}{n}+\binom{nh}{ 2}\frac{1}{n^2}+..........\right]$

$\displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]$

$\displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{n^2h^2-nh}{2!}\;\frac{1}{n^2}+......\right]$

$\displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{h^2}{2!}-\frac{nh}{n^2}+....\right]$

and all the terms with higher power of "n" in the denominator positions go to zero.
• December 6th 2010, 03:49 PM
alexandros
Quote:

$h\rightarrow\ 0$

$\displaystyle\ e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\rig ht)^n$

$\displaystyle\ e^h=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\r ight)^{nh}$

.

For that you assume the following theorem:

If ........... $lim_{n\to\infty(x_{n})=x\Longrightarrow lim_{n\to\infty}(x_{n})^h=x^h$
Where h is a real No

BUT is there such a theorem
• December 6th 2010, 04:02 PM
Quote:

Originally Posted by alexandros
.

For that you assume the following theorem:

If ........... $lim_{n\to\infty(x_{n})=x\Longrightarrow lim_{n\to\infty}(x_{n})^h=x^h$
Where h is a real No

BUT is there such a theorem

Please research Newton's Generalised Binomial theorem.
• December 6th 2010, 06:47 PM
alexandros
Quote:

Please research Newton's Generalised Binomial theorem.

It IS NOT a question of Newton,s Generalised Binomial, where the exponent is always natural.

It is a question whether the theorem i mention holds or not ,so that step 2 in your proof can be justified
• December 6th 2010, 08:32 PM
Drexel28
Quote:

Originally Posted by alexandros
It IS NOT a question of Newton,s Generalised Binomial, where the exponent is always natural.

It is a question whether the theorem i mention holds or not ,so that step 2 in your proof can be justified

If you're asking whether $x_n\to x\implies x_n^\alpha\to x^\alpha$ the answer is yes, by continuity of $f(y)=y^\alpha$.
• December 7th 2010, 01:56 AM
Quote:

Originally Posted by alexandros
It IS NOT a question of Newton,s Generalised Binomial, where the exponent is always natural. ????

It is a question whether the theorem i mention holds or not ,so that step 2 in your proof can be justified

Why can't you do some research???

Newton generalised the Binomial Theorem for real numbers a long time ago!

Also..

$e=2.71828.....\Rightarrow\ e^h=(2.71828....)^h$

$\displaystyle\ e=\left[\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n\right]\Rightarrow\ e^h=\left[\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n\right]^h$

$a=b\Rightarrow\ a^h=b^h$
• December 11th 2010, 06:39 PM
alexandros
Quote:

$h\rightarrow\ 0$

$\displaystyle\ e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\rig ht)^n$

$\displaystyle\ e^h=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\r ight)^{nh}$

$\displaystyle\ =\lim_{n\rightarrow\infty}\left[\binom{nh}{0}+\binom{nh}{1}\frac{1}{n}+\binom{nh}{ 2}\frac{1}{n^2}+..........\right]$

$\displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]$

$\displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{n^2h^2-nh}{2!}\;\frac{1}{n^2}+......\right]$

$\displaystyle\ =\lim_{n\rightarrow\infty}\left[1+h+\frac{h^2}{2!}-\frac{nh}{n^2}+....\right]$

and all the terms with higher power of "n" in the denominator positions go to zero.

So we have that: $\left(1+\frac{1}{n}\right)^{nh}$ =
$\left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]$

Does this equation work for n=1 and h= 1/2 ?
• December 12th 2010, 03:36 AM
Quote:

Originally Posted by alexandros
So we have that: $\left(1+\frac{1}{n}\right)^{nh}$ =
$\left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]$

Does this equation work for n=1 and h= 1/2 ?

What did the research indicate ?

You could compare it to...

$\displaystyle\sqrt{2}=\sum_{k=0}^{\infty}(-1)^{k+1}\frac{(2k-3)!!}{(2k)!!}=1+\frac{1}{2}-\frac{1}{2(4)}+\frac{3}{2(4)6}-\frac{3(5)}{2(4)6(8)}+.....$

What does

$\displaystyle\ 1+h+\frac{(nh)(nh-1)}{2!}+\frac{nh(nh-1)(nh-2)}{3!}+\frac{nh(nh-1)(nh-2)(nh-3)}{4!}+.....$

equal for those values of n and h ?
• December 12th 2010, 05:52 PM
alexandros
Can research show us that:

$\sqrt 2 = \frac{3}{2}$??
• December 12th 2010, 09:24 PM
chisigma
Quote:

Originally Posted by pflo
One way determine the deriveative of $e^x$ is to use inverse functions (natural logs). Another way is to use the definition of a derivative. Thus, $\frac{d}{dx}e^x=\lim_{h\rightarrow0}\frac{e^{x+h}-e^x}{h}$. When simplified, this results in the equation $\frac{d}{dx}e^x=e^x \lim_{h\rightarrow0}\frac{e^h-1}{h}$.

My questions is: what is a proof that $\lim_{h\rightarrow0}\frac{e^h-1}{h}=1$?

Your question is fully justified!.. effectively the limit $\displaystyle \lim_{x \rightarrow 0}\frac{e^x-1}{x}=1$ is a 'fundamental limit' from which You can derive all the properties of the exponential function. To demonstrate it we can start from the well known limit...

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

Now if $x \ne 0$ You can set $\displaystyle e^{x} = 1 + \frac{1}{r}$ so that is $\displaystyle x= \ln (1+\frac{1}{r})$ and therefore...

$\displaystyle \lim_{x \rightarrow 0} \frac{e^{x}-1}{x} = \lim_{r \rightarrow \infty} \frac{\frac{1}{r}}{\ln (1+\frac{1}{r})} = \lim_{r \rightarrow \infty} \frac{1}{\ln (1+\frac{1}{r})^{r}} = \frac{1}{\ln e} = 1$ (2)

http://digilander.libero.it/luposaba...ato&#91;1].jpg

Merry Christmas from Italy

$\chi$ $\sigma$
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