TITLE: A limit of a function with two different $n^{\text{th}}$ roots
QUESTION [7 upvotes]: $$\lim_{x \to \infty} \root 3 \of {x^3+3x^2} -\root 2 \of {x^2-2x}$$
In the problem above, I thought that having $$[\infty - \infty ]$$ transformed into $$\biggl [ {0\over {0}} \biggl]$$ and the using the L'Hopital's rule would solve it, but I ended up with a lot of confusing fractions of roots to different powers. I calculated a bit into it, but the deeper I went the more complicated it would get.
So my question is: Is there any other approach to this problem?

REPLY [9 votes]: We will use the expansion
$$
(1+u)^\alpha=1+\alpha u+O(u^2)
$$
as $u\rightarrow 0$.
Now 
$$
\sqrt[3]{x^3+3x^2}=x(1+3/x)^{1/3}=x(1+1/x+O(1/x^2))=x+1+O(1/x)
$$
and 
$$
\sqrt{x^2-2x}=x(1-2/x)^{1/2}=x(1-1/x+O(1/x^2))=x-1+O(1/x)
$$
as $x\rightarrow +\infty$.
So subtraction yields
$$
\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}=2+O(1/x) \rightarrow 2
$$
as $x\rightarrow +\infty$.
So your limit is $2$.

REPLY [3 votes]: Several of the other answers have relied on power series, which may or may not be far ahead of where you are in your studies.  It is a powerful technique that provides a great deal insight, but it isn't much use if you've only started looking at derivatives.
It sounds like you've practiced a lot of other limit problems already.  Likely you have seen a way to compute limits with a single root like $\lim_{x\to\infty} \sqrt{x^2-2x} - x$.  If not, then it is a standard trick to use the identity $$a-b = \frac{a^2 - b^2}{a+b},$$
and similarly for cube roots one can use
$$a-b = \frac{a^3-b^3}{a^2+ab+b}.$$
A simple way to deal with the two different roots here is to split the expression as
$$\lim_{x\to\infty} (\sqrt[3]{x^3+3x^2} - x) - (\sqrt{x^2-2x} - x).$$
We're using the fact that both roots are very close to $x$ and so it's reasonable to hope that each of the separate limits exists (in which case we can evaluate the limit of the difference).  In fact the first limit does converge to $1$ and the second converges to $-1$.