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$$\lim_{x \to \infty} \root 3 \of {x^3+3x^2} -\root 2 \of {x^2-2x}$$

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?

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2  
Did you mean $x \to \infty$ in the limit? –  copper.hat Jan 6 '13 at 18:30
    
Yes, absolutely. Thanks for noticing :) –  TerryGoldenstein Jan 6 '13 at 18:35

4 Answers 4

up vote 9 down vote accepted

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 $2$.

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(+1) nice technique. –  Mhenni Benghorbal Jan 6 '13 at 20:00

Hint: Are they really different roots? One of them is $\sqrt[6]{(x^3+3x^2)^2}$ and the other is $\sqrt[6]{(x^2-2x)^3}$.

Then use the identity $a^6-b^6=(a-b)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5)$.

Remark: It helps to first find the answer informally. The idea is that $\sqrt[3]{x^3+3x^2}$ will be roughly $x+1$, and $\sqrt[2]{x^2-2x}$ will be roughly $x-1$, so the difference, for large $x$, should be $2$-ish.

This approach can be made more formal, and is to my mind more satisfactory than the algebraic manipulations that are often expected, and that the hint described.

The first expression is $x\sqrt[3]{1+\frac{3}{x}}$. Find the first three terms of the power series expansion of $\sqrt[3]{1+\frac{3}{x}}$.

Similarly, the second expression is $x\sqrt[2]{1-\frac{2}{x}}$. Find the first three terms in the power series expansion of $\sqrt[2]{1-\frac{2}{x}}$. The answer will fall out.

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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 $-1$.

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You can also pull the dominant power out of each root and perform a Taylor expansion on the resulting root expressions:

$$ (x^3+3 x^2)^{1/3} - (x^2-2 x)^{1/2} = x [ (1+3/x)^{1/3} - (1-2/x)^{1/2} ] $$

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