I need to find the lim (squ(1+2x)-squ(1+3x)) / (x+2x^2) as x goes to 0 and where x>0.
Would you multiply by the conjugate?
$\displaystyle \lim_{x \to 0} \frac{\sqrt{1 + 2x} - \sqrt{1 + 3x}}{x + 2x^2} = \lim_{x \to 0} \frac{(\sqrt{1 + 2x} - \sqrt{1 + 3x})}{(x + 2x^2)} \cdot \frac{ (\sqrt{1 + 2x} + \sqrt{1 + 3x}) }{(\sqrt{1 + 2x} + \sqrt{1 + 3x}) }$
$\displaystyle = \lim_{x \to 0} \frac{-x}{(x + 2x^2) (\sqrt{1 + 2x} + \sqrt{1 + 3x}) } = ....$