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. 
 A: OK, without l'Hôpital... do $x \mapsto u^6$:
$$
\begin{align*}
\lim_{x \to 1} \frac{1}{2 (1 - \sqrt{x})} - \frac{1}{3 (1 - \sqrt[3]{x})}
  &= \lim_{u \to 1} \frac{1}{2 (1 - u^3)} - \frac{1}{3 (1 - u^2)} \\
  &= \lim_{u \to 1} 
      \frac{3 (1 - u^2) - 2 (1 - u^3)}{6 (1 - u^2)(1 - u^3)} \\
  &= \lim_{u \to 1}
        \frac{1 - 3 u^2 + 2 u^3}{6 (1 - u)^2 (1 + u) (1 + u + u^2)} \\
  &= \lim_{u \to 1}
        \frac{(1 - u)^2 (1 + 2 u)}{6 (1 - u)^2 (1 + u) (1 + u + u^2)} \\
  &= \lim_{u \to 1} \frac{1 + 2 u}{6 (1 + u) (1 + u + u^2)} \\
  &= \frac{1}{12}
\end{align*}
$$
A: 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
$$\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}=\frac{1}{h+\frac{1}{4}h^2+o(h^2)}-\frac{1}{h+\frac{1}{3}h^2+o(h^2)}\sim_0 \frac{\frac{1}{3}h^2-\frac{1}{4}h^2}{h^2}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$$
