Proof - the derivative of a scalar multiple

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hayesk85
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Homework Statement



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.

Start with: f(x) = k g(x) End: f'(x) = k g'(x)



Homework 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)

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)
 
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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
 
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hayesk85 said:

Homework Statement



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.

Start with: f(x) = k g(x) End: f'(x) = k g'(x)



Homework 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)

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

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