Find the derivative of $ f(x) = x^9 - x^7$ using limit definition I can find this without using the limit definition (I think the formula is 
$\frac{f(x+h) - f(x)}{h}$  
My first steps to solving are $f'(x) \lim\limits_{h\rightarrow 0} \frac{(x+h)^9 - (x+h)^7 - x^9 + x^7}{h}$
Thank you in advanced! This problem is making my head throb... and I already know the derivative is $9x^8 - 7x^6$
EDIT: Thank you everyone for the extra push! I was in the right direction, I just overcomplicated and did not realize that I could factor the h out of the problem after using the binomial theorem. 
After plugging in the lim h->0   I had f'(x) - x^9+9x^8-7x^6-x^7-x^9+x^7 = 9x^8 - 7x^6
 A: HINT:
$$(x+h)^9 - (x+h)^7 - x^9 + x^7$$
$$=\binom91x^8h+\binom92x^7h^2+O(h^3)-\left[\binom71x^6h+\binom72x^5h^2+O(h^3)\right]$$
As $h\to0,h\ne0$ so cancel $h$ safely.
In fact using Binomial series, 
$(x+h)^n=x^n\left(1+\dfrac hx\right)^n=x^n\left(1+n\cdot\dfrac hx+O(h^2)\right)=x^n+nx^{n-1}h+O(h^2)$
$$\implies\dfrac{d(x^n)}{dx}=\lim_{h\to0}\dfrac{(x+h)^n-x^n}h=\lim_{h\to0}\dfrac{nx^{n-1}h+O(h^2)}h=nx^{n-1}$$
A: You can also use
$\displaystyle f^{\prime}(x)=\lim_{t\to x}\frac{f(t)-f(x)}{t-x}=\lim_{t\to x}\frac{(t^9-t^7)-(x^9-x^7)}{t-x}=\lim_{t\to x}\frac{(t^9-x^9)-(t^7-x^7)}{t-x}$
$=\displaystyle\lim_{t\to x}\frac{(t-x)[t^8+t^7 x+t^6 x^2+\cdots+x^8]-(t-x)[t^6+t^5 x+t^4 x^2+\cdots+x^6]}{t-x}$
$=\displaystyle\lim_{t\to x}\left([t^8+t^7 x+t^6 x^2+\cdots+x^8]-[t^6+t^5 x+t^4 x^2+\cdots+x^6]\right)=9x^8-7x^6$
A: Good start.
From here, you can expand the binomials, such as $(x+h)^9$ without as much effort as you might think, by using the binomial theorem.
See if you can do that, then start factoring the numerator.
