Question

I was proving d/dx (e^x). Our teacher mentioned the proof, but I have a question.
Anyone ?

Answer 1

\[\LARGE \lim_{h \rightarrow 0}~\frac{f(x+h)-f(x)}{h}\]\[\LARGE \lim_{h \rightarrow 0}~\frac{e^{x+h}-e^{x}}{h}\]\[\LARGE \lim_{h \rightarrow 0}~\frac{e^{x}e^{h}-e^{x}}{h}\]\[\LARGE \lim_{h \rightarrow 0}~\frac{e^{x}(e^{h}-1)}{h}\]\[\LARGE e^{x} \times \lim_{h \rightarrow 0}~\frac{(e^{h}-1)}{h}\]

Answer 2

Why is the last limit,
\[\Large \color{red}{\lim_{h \rightarrow 0}~\frac{(e^{h}-1)}{h}}\]
equal to 1 ?

Answer 3

what is e^0 - 1?

Answer 4

yes, so that is just saying that 0/0 = 1 ?

Answer 5

I mean e^0-1 = 1-1 =0

So the limit is 0/0, and that is 1. That's what you want to say ?

Answer 6

use LH rule

Answer 7

Haha, I haven't proved e^x yet

Answer 8

I won't be able to know e^h like this, this will become an endless cycle, see ?

Answer 9

thats what i was leaning towards yes, 0/0 limits to 1

Answer 10

I was always thinking that it 0/0 is undefined. But isn't there anything better to this than saying that 0/0=1 ?

Answer 11

no :O

Answer 12

I looked up on yahoo, but all I can see is L'H's. Well, I am familiar with it, despite that I'm only taking Calc1, but even using L'H's I am unable to get the limit, because although when I derive the bottom, the "h" (w/ respect to h) I get 1, I will have to derive e^h on the top and that is not proven yet (for me) to do this.

Answer 13

well the limit is 0/0
thus ur allowed to use LH rule

when limit f(x)/g(x) as x goes to 0 is 0/0
then
limit f(x)/g(x) as x goes to 0=limit f'(x)/g'(x) as x goes to 0

Answer 14

No, L'H's doesn't work for my given (above) reason.

Answer 15

Math sucks !

Answer 16

(apparently)

Answer 17

e^h = 1 + h + h^2/2! + h^3/3! + ....

e^h - 1 = h + h^2/2! + h^3/3! + ....

(e^h-1)/h = 1 + h/2! + h^2/3! + ...

when h=0, 1+ 0 + 0 + 0 + ....

Answer 18

ooh, cool. I kind of heard that, but I never thought that there will be a need for that. It so nicely intervenes over here!
Tnx armistre!

Answer 19

youre welcome :)

Answer 20

Sorry, can't write a second testimonial -:(

Answer 21

The series expansion also uses the derivative of e^x for the coefficients, right? Or can it come through other means that I just didn't think of? :P

Answer 22

well, that is one way to develop the series yes. but if its already known before hand

Answer 23

well, the coefficient if you mean something like d/dx e^(2x)


lim f(x+h) - f(x)
h approaches 0 ------------
h


lim e^(2(x+h)) - e^(2x)
h approaches 0 ------------
h


lim e^(2x)e^(2h) - e^(2x)
h approaches 0 ------------------
h

lim e^(2x) (1 - e^(2h) )
h approaches 0 ------------------
h


lim (1 - e^(2h) )
e^(2x) times h approaches 0 -------------
h


e^(2h) - 1^2 = h^2 +....


e^(2h) - 1 = h^2 +....


when h=0, 1+ 0 + 0 + 0 + ....

Answer 24

Sorry, I wanted to post earlier but OS bonked out and I had to go.

Got it, although I guess I just don't find it convincing enough personally!

One thing that I was thinking about was implicit differentiation by taking the natural log. of both sides. But I figured maybe you wanted the limit definition proof specifically, or the natural log derivative was also off-limits.

Here was something I stumbled across that looked promising:
https://www.physicsforums.com/threads/solving-for-derivative-of-e-x-using-limit-definition.161965/#post-1280952
It is a bit more involved but the actions seem sound. Basically just using the substitutions of variables and definition of e to make the simplifications necessary.

Answer 25

And no, by coefficients I meant the coefficients of the expanded series form of e^x.
\( \displaystyle e^x = \sum_{n=0}^{\infty} \color{green}{\dfrac{1}{n!} } x^n\)

You can derive the coefficients by repeated differentiation.
e^x = a_0 + a_1 x + a_2 x^2 + ... x=0
e^0 = a_0 + 0 + 0 + ... = a_0 = 1
d/dx(e^x) = d/dx(a_0 + a_1 x + a_2 x^2 + ... )
e^x = a_1 + 2a_2 x + .3a_3 x^2 + ... x=0
e^0 = a_1 + 0 + 0 + 0... = a_1 = 1.
Due to power rule, you can see that the next will be 2a_2 = 1, or 1/2 = 1/2!. Then 3*2 a_3 = 1, or 1/3/2 = 1/3!. And so on. The only problem is that you're using derivatives of e^x again, unless you already have this form given to you.

Answer 26

the proof of derivatives of lnx and e^x involve some given limits related to definition of e
and uses the ln inverse of e^x to obtain the result
without series expansion

Answer 27

you may want to check MIT videos to see how they did it
i think the prof that is lecturing proved it

Answer 28

you can prove it using implicit differentiation
y=e^x
x=lny
given we know d/dx(lnx)
you can prove it easily using the definition