Calculus Bonus Problem solve algebraically
$$\lim_{x\rightarrow 0}\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x+x^2}$$
without using L'hopital's rule
This is a bonus question for my Math Intensive Major Cal 1 class, I would like to know how to solve it since it stumped me on the test, and after many hours of working at this problem it has stumped me again and again

We will use the formula
$$
a-b=\frac{a^{12}-b^{12}}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}
$$
with 
$$
a=\sqrt[3]{1+x^2}\quad\mbox{and}\quad b=\sqrt[4]{1-2x}.
$$
This gives
$$
\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}=\frac{(1+x^2)^4-(1-2x)^3}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}=\frac{6x-8x^2+\ldots}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}
$$
$$
\frac{2x(3-4x+\ldots)}{a^{11}+a^{10}b+\ldots+ab^{10}+b^{11}}.
$$
So 
$$
\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x+x^2}=\frac{2(3-4x+\ldots)}{(1+x)(a^{11}+a^{10}b+\ldots+ab^{10}+b^{11})}.
$$
Now note that both $a$ and $b$ tend to $1$, so your limit is
$$
\frac{2\cdot 3}{12}=\frac{1}{2}.
$$
