Derivative of e^x by Limit Definition

Date: 05/21/2002 at 22:40:18
From: Jeff King
Subject: derivative of e^x by limit definition

Dear Dr. Math,

I've tried (as has my entire calculus class) to prove that the 
derivative of e^x is e^x by the limit definition of the derivative 
(without using the Taylor expansion for e^x) and we cannot seem to 
get past one last step.  

In setting up the form of the limit definition of the derivative we've 
made e(x+delta x) into e^x*e^(delta x) for convenience and left the 
-e^x in the numerator of the limit definition. Then by factoring we 
determined that the limit as delta x goes to 0 is 

  (e^x(e^(delta x)-1))/delta x.  

The e^x alone in the numerator would seem to be our best bet, but we 
can't make the rest of the function go to 1 as some people have 
guessed it should. 

How would one prove the limit definition or just that 

  (e^(delta x)-1)/(delta x) = 1 

as x tends to infinity?  Thanks for your assistance!

Date: 05/29/2002 at 22:41:49
From: Doctor Fwg
Subject: Re: derivative of e^x by limit definition

Dear Jeff,

Here is a possible solution that does not involve the use of a Taylor 
expansion.  I hope the notation I have used here will be easy enough 
to follow.

Definition of e is:

  1)  e = Limit   (1 + 1/n)^n
           n->Inf

However, if n = 1/h, then

  2)  e = Limit   (1 + h)^(1/h)
           h->0

If f(x) = e^x, the definition of f'(x) is:

  3)  f'(x) = Limit  [f(x + h) - f(x)]/h
               h->0

But:

  4)  f(x + h) = e^(x+h) = (e^x)(e^h)

So:

  5)  f'(x) = Limit [(e^x)(e^h) - e^x]/h 
               h->0

            = Limit [(e^x)((e^h) - 1)]/h
               h->0

But, raising both sides of equation (2) to the power of h yields:

  6)  e^h = Limit [(1 + h)^(1/h)]^h 
             h->0

          = Limit [(1+ h)]
             h->0

Placing this value of e^h into equation (5) yields:

  7)  f'(x) = Limit [(e^x){(1 + h) - 1}]/h 
               h->0

            = Limit [e^x(h/h)]
               h->0

But, h/h = 1, so:

  8)  f'(x) = Limit [e^x(1)] 
               h->0

            = e^x

With Best Regards,

Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/