1. ## limit proof

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

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

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

My question is: what is a proof that $\displaystyle \lim_{h\rightarrow0}\frac{e^h-1}{h}=1$?
$\displaystyle \displaystyle\ e^h=1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+.......$

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

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

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

3. Originally Posted by Archie Meade
$\displaystyle \displaystyle\ e^h=1+h+\frac{h^2}{2!}+\frac{h^3}{3!}+.......$

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

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

then take the limit as $\displaystyle h\rightarrow\ 0$
To get the above you use the Taylor's expansion where :

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

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

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

But that is what you want to prove

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

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

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

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

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

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

we obtain

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

from which we can write the power series expansion of $\displaystyle e^h$

5. Is h a real or a natural No??

6. Originally Posted by alexandros
Is h a real or a natural No??
$\displaystyle h\rightarrow\ 0$

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

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

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

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

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

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

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

7. Originally Posted by Archie Meade
$\displaystyle h\rightarrow\ 0$

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

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

For that you assume the following theorem:

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

BUT is there such a theorem

8. Originally Posted by alexandros
.

For that you assume the following theorem:

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

BUT is there such a theorem
Please research Newton's Generalised Binomial theorem.

9. Originally Posted by Archie Meade
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

10. 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 $\displaystyle x_n\to x\implies x_n^\alpha\to x^\alpha$ the answer is yes, by continuity of $\displaystyle f(y)=y^\alpha$.

11. 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..

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

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

$\displaystyle a=b\Rightarrow\ a^h=b^h$

12. Originally Posted by Archie Meade
$\displaystyle h\rightarrow\ 0$

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

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

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

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

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

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

and all the terms with higher power of "n" in the denominator positions go to zero.
So we have that:$\displaystyle \left(1+\frac{1}{n}\right)^{nh}$ =
$\displaystyle \left[1+h+\frac{(nh)(nh-1)}{2!}\;\frac{1}{n^2}+........\right]$

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

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

Does this equation work for n=1 and h= 1/2 ?
What did the research indicate ?

You could compare it to...

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

What does

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

equal for those values of n and h ?

14. Can research show us that:

$\displaystyle \sqrt 2 = \frac{3}{2}$??

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

My questions is: what is a proof that $\displaystyle \lim_{h\rightarrow0}\frac{e^h-1}{h}=1$?
Your question is fully justified!.. effectively the limit $\displaystyle \displaystyle \lim_{x \rightarrow 0}\frac{e^x-1}{x}=1$ 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...

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

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

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

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

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