<s>[INST] Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$ 
Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$

I would like a little help, so that I can finish solving this exercise.
So far, I've got
$$
\frac{\frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{h}
= \frac{\frac{(x-a)^2 - (x-a-h)^2}{(x-a-h)^2(x-a)^2}}{ \frac{h}{1}}
= \frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}
$$
In the numerator maybe I can continue with a difference of squares, but I'm a little confused.
 [/INST] $$\frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}=\frac{\left((x-a)-(x-a-h)\right)\left((x-a)+(x-a-h)\right)}{h(x-a-h)^2(x-a)^2}=$$
$$=\frac{h(2x-2a-h)}{h(x-a-h)^2(x-a)^2}=\frac{(2x-2a-h)}{(x-a-h)^2(x-a)^2}$$
If now you want to take the limit when $\;h\to0\;$ , for example, the above equals 
$$\frac2{(x-a)^3}$$
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