Any 'proof' needs to somehow introduce ln(a) in a non-artificial way.  Using derivatives, as Alex points out, is not the way.

However, we know that .  If we expand that exp into a series to 2 terms, we get .  Now we're nearly there!  Plug that in to the limit expression and we get

, which is a valid computation as h>0.

 because we cannot see the exponent that you want to expand.

in both case, you have to prove:

> deq1 := exp(x)*(Limit((exp(h)-1)/h, h = 0));

                                   /       /exp(h) - 1\\
                    deq1 := exp(x) | lim   |----------||
                                   \h -> 0 \    h     //

The liimit is 1.  But you have the result by the Hospital's Rule, wich you cannot do since it's what you are trying to prove.

> deq2 := a^x*(Limit((a^h-1)/h, h = 0));

                                   /       / h    \\
                                 x |       |a  - 1||
                        deq2 := a  | lim   |------||
                                   \h -> 0 \  h   //

While the limit is ln(a) by again using the Hospital's Rule.

For the moment, all I have to illustrate that the limit is ln(a) is to use the spreadsheet and show that the limit tend to be, for a = 5 for example, ln(5).  Of course, I would like a precises theoriticall.

What you will accept as a proof depends on what you already know.

As has already been pointed out, using l'Hopital's Rule is suspect because that presumes knowledge of the derivative of a^x, or exp(x). There are ways to get this using other facts, e.g., series and inverse function. But, this brings in more questions about what is proven and what is accepted fact for working problems.

Can you provide us with some additional background for your question?

I assume it's for an assignment you have been given. For what course? What do you know from previous courses? Can you provide any similar problems (or examples) that would illuminate us on the general level of your course?

Lastly,  your initial post asked for a "proof using Maple". What does this mean, exactly? Do you simply want to verify the calculations at each step? Or, might a "graphical proof" be acceptable?

Doug

---------------------------------------------------------------------
Douglas B. Meade  <><
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.ed

My goal, in teaching a first course of calculus, is to be able to calculate the derivative of the function a^x ,  ln(x) and sin(x)  with the definition:

> Diff(f(x), x) = Limit((f(x+h)-f(x))/h, h = 0);

                      d                /f(x + h) - f(x)\
                     --- f(x) =  lim   |---------------|
                      dx        h -> 0 \       h       /

First, I show that:

> Limit((1+h)^(1/h), h = 0) = e;

                                          /1\    
                                          |-|    
                                          \h/    
                             lim   (1 + h)    = e
                            h -> 0               

with a list that give the value of the limit as h approche 0.  I don't know any other way to demonstrate.

Then I look for a^x

> deq1 := a^x*(Limit((a^h-1)/h, h = 0));

                                   /       / h    \\
                                 x |       |a  - 1||
                        deq1 := a  | lim   |------||
                                   \h -> 0 \  h   //

I show that for a = 2, the limit is 0.6931 and that for a = 3, the limit is 1.0986.  So is there a function that the derivative is the function of herselft.  So we look at:

> Limit((a^h-1)/h, h = 0) = 1;

                                    / h    \    
                                    |a  - 1|    
                              lim   |------| = 1
                             h -> 0 \  h   /    
           

and the solution is a = e.  But for  a different value of a, I am stuck with:

> deq2 := a^x*(Limit((a^h-1)/h, h = 0));

                                   /       / h    \\
                                 x |       |a  - 1||
                        deq2 := a  | lim   |------||
                                   \h -> 0 \  h   //

Then I go with :

a = 2 in deq2 give 0.6931471806*2^x = ln(2)*2^x

a = 3 in deq2 give 1.098612289*3^x = ln(3)*3^x

And so on and I deduct that the result is in general:  I am not very please with that way of proving this.  So for  now, I have show that

> Diff(a^x, x); % = value(%);

                               d   x    x      
                              --- a  = a  ln(a)
                               dx              

Of course, I could use the following

> Diff(ln(a^x), x) = (Diff(a^x, x))/a^x;

                                           d   x
                                          --- a 
                              d    / x\    dx   
                             --- ln\a / = ------
                              dx             x  
                                            a   
> Diff(x*ln(a), x) = ln(a);

                             d                   
                            --- (x ln(a)) = ln(a)
                             dx                  

But to do that, I need to know the derivative of ln(x)

> Diff(Log[a](x), x) = Limit((log[a](x+h)-log[a](x))/h, h = 0);

                                        /ln(x + h)   ln(x)\
                                        |--------- - -----|
                  d                     |  ln(a)     ln(a)|
                 --- Log[a](x) =  lim   |-----------------|
                  dx             h -> 0 \        h        /


                                                   /  /x + h\\
                                                   |ln|-----||
                      /ln(x + h)   ln(x)\          |  \  x  /|
                      |--------- - -----|    lim   |---------|
                      |  ln(a)     ln(a)|   h -> 0 \    h    /
                lim   |-----------------| = ------------------
               h -> 0 \        h        /         ln(a)
and       

                                  /  /x + h\\    
                                  |ln|-----||    
                                  |  \  x  /|   1
                            lim   |---------| = -
                           h -> 0 \    h    /   x

wich I have to prove again.  I have a nice way for doing so:

 `Δy`/`Δx` = Log[a](1+`Δx`/x)/`Δx`;

                                        /    Delta;x\
                                  Log[a]|1 + --------|
                       Delta;y         \       x    /
                       -------- = --------------------
                       Delta;x         Δx      

Multiplying and dividing the last output and setting Delta;x/x

= alpha we have

> `Δy`/`Δx` = Log[a](1+alpha)^(1/alpha)/x;

                                                 /  1  \
                                                 |-----|
                                                 \alpha/
                     Δy   Log[a](1 + alpha)       
                     -------- = ------------------------
                     Δx              x            
and having prove (with a table of numbers) that

                                          /1\    
                                          |-|    
                                          \h/    
                             lim   (1 + h)    = e
                            h -> 0               

then

                           d              Log[a](e)
                          --- Log[a](x) = ---------
                           dx                 x    

And I prove the same way that

> Diff(sin(x), x) = cos(x)*(Limit(sin(h)/h, h = 0));

                     d                  /       /sin(h)\\
                    --- sin(x) = cos(x) | lim   |------||
                     dx                 \h -> 0 \  h   //

and that, with a Table;

                                    /sin(h)\    
                              lim   |------| = 1
                             h -> 0 \  h   /    

So I want to find the derivative of thoses 3 functions knowing only the derivative of x^n, the sum (minus) rule, the multipication rule, the quotient rule and the chain rule.

Is doing so, only with table of numbers, am I prooving something?  Does anyone have a beautyfull way to go?

Thanks in advance

 All I want to do is construct the rule of differenting the functions all with the formula

> Diff(f(x), x) = Limit((f(x+h)-f(x))/h, h = 0);

                      d                /f(x + h) - f(x)\
                     --- f(x) =  lim   |---------------|
                      dx        h -> 0 \       h       /

So I find the rule for x^n. Then I establish the sum (minus) rule, the multipication rule, the quotient rule.

Then I show that the definition of "e" is

> Limit((1+h)^(1/h), h = 0) = e;

                                          /1\    
                                          |-|    
                                          \h/    
                             lim   (1 + h)    = e
                            h -> 0               

Then knowing that:

y = a^x and log[a](y) = x

find the derivative of both function with the definitions that I have establish and with the formula of the beginning.

As for sin(x), the student know that it's a periodic function F(x) = F(x + 2*Pi).  And then we construct the rule of derivative for cos(x), tan(x) and so on....

And finally the chain rule and l'Hospital's rule.  So now they have all they need to find the derivative of every kind of functions.

So you see that the derivative of a^x, ln(x) and sin(x) are the functions that I want to prove without using table of the fucntion while h tend to zero and I cannot use a series because it implie that I know the derivative of the fuction we are looking for.

I hope this will help you.

When I studied that in high school, it was done as follows. First, few plots y=2^x, y=3^x, y=4^x etc. were plotted, together with their tangents at 0, and one could see visually that the slope, i.e. the derivative of y=2^x at 0 is less than 1, and the derivative of y=4^x at 0 is greater than 1. Then, one can prove that if the derivative of y=b^x (with b>0) at 0 is c, then the derivative of y=b^(kx) at 0 is kc (by substituting kh instead of h in the limit for the derivative of y=b^x at 0.) 

If c<>0. for k=1/c that gives the derivative 1 at 0. That can be used as a definition of e as the unique positive number such that the derivative of e^x at 0 is 1.

Now, for b=e, the derivative of e^(kx) at 0 is k, and since a=e^(ln a), that gives the value ln(a) for the derivative of y=a^x at 0.

Alec