Simplify $\frac{1}{h}\Big( \frac{1}{(x-a-h)^2} - \frac{1}{(x-a)^2}\Big)$

Question

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.

Answer 1

$$\frac{(x-a)^2 - (x-a-h)^2}{h(x-a-h)^2(x-a)^2}=\frac{\left((x-a)-(x-a-h)\right)\left((x-a)+(x-a-h)\right)}{h(x-a-h)^2(x-a)^2}=$$

$$=\frac{h(2x-2a-h)}{h(x-a-h)^2(x-a)^2}=\frac{(2x-2a-h)}{(x-a-h)^2(x-a)^2}$$

If now you want to take the limit when $\;h\to0\;$ , for example, the above equals

$$\frac2{(x-a)^3}$$

Answer 2

The top simplifies $$ (x-a)^2 -[(x-a)-h]^2 =$$

$$(x-a)^2 -[(x-a)^2-2h(x-a)+h^2]$$

$$=2h(x-a)-h^2$$

Answer 3

hint

Begin by putting $$x-a=b$$ then

$$\frac{1}{(b-h)^2}-\frac{1}{b^2}=$$ $$\frac{b^2-(b-h)^2}{b^2(b-h)^2}=$$

$$\frac{h(2b-h)}{b^2(b-h)^2}.$$

And if we divide by h, one find

$$\frac{2(x-a)-h}{(x-a)^2(x-a-h)^2}.$$