Proof of the Power Rule - Stuck at the End

[Astrum]

Homework Statement

Prove that the following

$f'(x)=nx^{n-1}$ if $f(x)=x^{n}$

Homework Equations

Binomial theorem, definition of the derivative

The Attempt at a Solution

$$f'(x)=lim_{h\rightarrow0}\frac{f((x+h)^{n})-f(x)}{h}$$

We need to expand the (x+h)^2 term now

$$\sum^{n}_{k=0}{n\choose k} x^{n-k}h^{k}={n\choose 0}x^n+{n\choose 1}x^{n-1}h+...+{n\choose k}x^{n-k}h^{k}$$

So, we sub this for the f((x+h)^n) term:

$$lim_{h \rightarrow 0}\frac{{n\choose 0}x^n+{n\choose 1}x^{n-1}h+...+{n\choose k}x^{n-k}h^{k}-x}{h}$$

$${n\choose k}=\frac{n!}{(n-k)!k!}$$

This now simplifies to---

$$lim_{h\rightarrow0}\frac{x^{n}}{h}+nx^{n-1}-lim_{h\rightarrow0}\frac{x}{h}$$

The first and third terms create only a one sided limit, and they both go to infinity, I'm not sure where I went wrong...

I could just look up the proof, but I'm trying to do it by myself, so I'm only looking for a hint.

I feel like I'm really close, because I have the answer there, I just don't know why those other two terms are messing it up.

There are only two possibilities, either I've gone in the completely wrong direction, or I made a silly mistake somewhere.

 

[SteamKing]

Your definition of f'(x) is a little strange.

Since f(x) = x^n, then

f'(x) = lim (h -> 0) [(x+h)^n - x^n) / h]

Try expanding now using the Binomial Theorem.

 

[Astrum]

Your definition of f'(x) is a little strange.

Since f(x) = x^n, then

f'(x) = lim (h -> 0) [(x+h)^n - x^n) / h]

Try expanding now using the Binomial Theorem.

Thanks, I didn't catch that... how silly.

So, that means those two terms cancel out, and we're left with the correct one.

 

[LCKurtz]

Your definition of f'(x) is a little strange.

Since f(x) = x^n, then

f'(x) = lim (h -> 0) [(x+h)^n - x^n) / h]

Try expanding now using the Binomial Theorem.

Thanks, I didn't catch that... how silly.

So, that means those two terms cancel out, and we're left with the correct one.

What do you mean by "we're left with the correct one"? There are two terms that do cancel but there are a lot of others and you have to deal with the h's. Have you actually worked it out?

 

[Astrum]

What do you mean by "we're left with the correct one"? There are two terms that do cancel but there are a lot of others and you have to deal with the h's. Have you actually worked it out?

Yes, I skipped steps out of laziness (meaning, I did them on paper, but I didn't feel like taking the time to write them out here).

The h on the bottom cancels out and all the hs on the top go to 0.