Determine the definite limit The following limit 
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
evaluates to 1/12. 
This is my progress so far:
$$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$
$$\lim_{x\to 1}\frac{1 + \sqrt{x}}{2(1 - x)} - \frac{1 + \sqrt[3]{x} + \sqrt[3]{x^2}}{3(1 - x)}$$
$$\lim_{x\to 1}\frac{3(1 + \sqrt{x})- 2(1 + \sqrt[3]{x} + \sqrt[3]{x^2})}{6(1 - x)}$$
And that's as far as I go. 

If we pose $x=1-h$ we find
$$\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}=\frac{1}{2(1 - \sqrt{1-h})} - \frac{1}{3(1 - \sqrt[3]{1-h})}$$
and we have $$\sqrt{1-h}=1-\frac{1}{2}h-\frac{1}{8}h^2+o(h^2)$$ and $$(1-h)^{1/3}=1-\frac{1}{3}h-\frac{1}{9}h^2+o(h^2)$$
hence we find
$$\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}=\frac{1}{h+\frac{1}{4}h^2+o(h^2)}-\frac{1}{h+\frac{1}{3}h^2+o(h^2)}\sim_0 \frac{\frac{1}{3}h^2-\frac{1}{4}h^2}{h^2}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$$
