Finding a derivative by definition

Are you allowed to break this apart into two limits?

You've got f(x) = g(x) + h(x), for g(x) = 1/sqrt[x] and h(x) = 2x^2. The h(x) limit is easy, and the g(x) limit isn't that hard, once it's pulled aside and you apply the conjugate (a trick you'll want to keep in mind for the next test):

$\frac{g(x\, +\, h)\, -\, g(x)}{h}\, =\, \frac{\frac{1}{\sqrt{x\, +\, h}}\, -\, \frac{1}{\sqrt{x}}}{h}$

$=\, \frac{\left(\frac{\sqrt{x}\, -\, \sqrt{x\, +\, h}}{\sqrt{x}\sqrt{x\, +\, h}}\right)}{h}\, =\, \frac{\sqrt{x}\, -\, \sqrt{x\, +\, h}}{\sqrt{x}\sqrt{x\, +\, h}(h)}$

$=\, \left(\frac{\sqrt{x}\, -\, \sqrt{x\, +\, h}}{\sqrt{x}\sqrt{x\, +\, h}(h)}\right)\left(\frac{\sqrt{x}\, +\, \sqrt{x\, +\, h}}{\sqrt{x}\, +\, \sqrt{x\, +\, h}}\right)$

$=\, \frac{x\, -\, (x\, +\, h)}{\sqrt{x}\sqrt{x\, +\, h}(h)\left(\sqrt{x}\, +\, \sqrt{x\, +\, h}\right)}$

$=\, \frac{-1}{\sqrt{x}\sqrt{x\, +\, h}\left(\sqrt{x}\, +\, \sqrt{x\, +\, h}\right)}$

Now take the limit. (Wink)