# Homework Help: Proof - the derivative of a scalar multiple

1. Sep 22, 2008

### hayesk85

1. The problem statement, all variables and given/known data

I am confused how the scalar multiple is divided out of the proof of this rule without taking an h with it in the denominator, which would get very tiny meaning the entire thing would go to infinity or negative infinity or zero, you can't tell.

2. Relevant equations

This is the proof I was given:

f'(x) = lim(h->0) [k g(x+h) - k g(x)] / h

f'(x) = lim(h->0) [k {g(x+h) - g(x)}] /h

Next step I do not agree with: (Never mind -this is legal, right?)
f'(x) = lim(h->0) k [{g(x+h) - g(x)}/h]

f'(x) = k lim(h->0) [{g(x+h) - g(x)}/h]

f'(x) = k g'(x)

3. The attempt at a solution

This is what I think would happen at the step I disagree with:

f'(x) = lim(h->0) k/h * [{g(x+h) - g(x)}/h]

f'(x) = lim(h->0) k/h * lim(h->0) [{g(x+h) - g(x)}/h]

f'(x) = ??? * g'(x)

Last edited: Sep 22, 2008
2. Sep 22, 2008

### Dick

(k/h)*(g(x+h)-g(x))/h=(k*g(x+h)-k*g(x))/h^2. That's not equal to (k*g(x+h)-k*g(x))/h. Why would you do that?

3. Sep 22, 2008

### jackiefrost

Check out the presentation of the formal definition of "limit" (the "delta-epsilon" formality):
http://mathforum.org/library/drmath/view/53403.html"

That doesn't answer your question, I know. But it's an unusually interesting discussion of what is really meant by "limit".

jf

Last edited by a moderator: Apr 23, 2017
4. Sep 22, 2008

### HallsofIvy

That's just bad algebra. You now have two "h" s in the denominators.