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
 A: 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
$$
\frac{2\cdot 3}{12}=\frac{1}{2}.
$$
A: As $x \rightarrow 0$, you can ignore the $x^2$ pieces, as they go to zero faster than $x$.  the limit is then equivalent to
$$\lim_{x \rightarrow 0} \frac{1-(1-2 x)^{1/4}}{x}$$
Now use the difference of two fourth powers:
$$y^4-z^4 = (y-z)(y+z)(y^2+z^2)$$
which means that
$$1-(1-2 x)^{1/4} = \frac{2 x}{(1+(1-2 x)^{1/4})(1+(1-2 x)^{1/2})}$$
The limit is then $2/((2)(2)) = 1/2$.
EDIT
Because a question was raised about the intuitive approach, I will make a little justification about my first statement.  Write
$$(1+x^2)^{1/3} - (1-2 x)^{1/4} =( 1-(1-2 x)^{1/4}) - (1-(1+x^2)^{1/3})$$
For the second term on the RHS, use $y^3-z^3 = (y-z)(y^2+yz+z^2)$ and get
$$1-(1+x^2)^{1/3} = \frac{-x^2}{1+(1+x^2)^{1/3} + (1+x^2)^{2/3}}$$
which, when divided by $x$, goes to zero in the limit.
A: 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.
