TITLE: Calculus Bonus Problem
QUESTION [5 upvotes]: 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

REPLY [3 votes]: 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
$$
\frac{2\cdot 3}{12}=\frac{1}{2}.
$$

REPLY [2 votes]: Although as comments above you are not familiar with the following fact; I am writing it here to keep it for the next.In fact @julien did the answer completely. The function $\alpha(x)$ can be said very small at $x\to a$ when $$\lim\alpha(x)\to 0$$ There are some facts about such this function. For example $\sqrt[n]{1+\alpha(x)}-1\sim\frac{\alpha(x)}{n}$. So $$\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}=(\sqrt[3]{1+x^2}-1)-(\sqrt[4]{1-2x}-1)~\sim~\frac{x^2}{3}-\frac{-2x}{4}=\frac{x^2}{3}+\frac{x}{2}$$ when $x\to 0$ from both sides. I think the problem gets easy from here.