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Firstly, this is not a homework. I just want to solve this limit for my own curiosity and self-learning. I have tried to solve this limit for 5-6 hours with no luck. Then I tried to read information about limits when there is case "inifinity - inifinity". Also I watched videos, but again with no luck. I think that this limit could be very difficult or very simple and my thinking is just wrong for finding it. So, it would be grateful if some of you could explain step by step solution of this limit. And there is the limit:

$$ \lim_{x\rightarrow +\infty} (\sqrt[3]{x^3+3x^2} - \sqrt{x^2-2x}) $$

What I did for now. It is sad to say, but almost nothing. All my efforts where just trying to factorize, simplify, remove denominator, etc. The "best" achieved result was more and more polynomials and just increasing level of difficulty to solve such type of problem.

I know that the answer for this limit is 2. I hope at least this will help.

Thank you for your any help in advance.

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I would reduce to expressions like $\sqrt[3]{1+\epsilon}$ and $\sqrt{1+\sigma}$ with $\epsilon \to 0$ and $\sigma \to 0$. But I think this would require tools you do not have, yet. –  Siminore Sep 28 '12 at 11:15

4 Answers 4

up vote 1 down vote accepted

Let $y=\frac{1}{x}$ which $$\implies \lim_{x\rightarrow +\infty} (\sqrt[3]{x^3+3x^2} - \sqrt{x^2-2x})=\lim_{y\to 0^+}\frac{((1+3y)^{1/3}-(1-2y)^{1/2})}{y}$$

Now you can use L'Hopital's rule to get $$\lim_{y\to 0^+}\frac{((1+3y)^{1/3}-(1-2y)^{1/2})}{y}=\lim_{y\to0^+}\frac{(1+3y)^{-2/3}+(1-2y)^{-1/2}}{1}=2$$

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You can write, as $x \to +\infty$, $$ \sqrt[3]{x^3+3x^2} = x \sqrt[3]{1+\frac{3}{x}}=x \left(1 + \frac{1}{x}-\frac{1}{x^2}+O(x^{-3})\right) =x+1-\frac{1}{x}+O(x^{-2}) $$ and $$ \sqrt{x^2-2x} = x \sqrt{1-\frac{2}{x}} = x \left( 1 - \frac{1}{x} - \frac{1}{2} \frac{1}{x^2}+O(x^{-3}) \right) = x-1-\frac{1}{2}\frac{1}{x}+O(x^{-2}). $$ Therefore $$ \sqrt[3]{x^3+3x^2} - \sqrt{x^2-2x} = 2 + o(1). $$

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$$\begin{align*} \lim_{x\to\infty}\left(\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\right)&=\lim_{x\to\infty}\left(x\left(1+\frac3x\right)^{1/3}-x\left(1-\frac2x\right)^{1/2}\right)\\ &=\lim_{x\to\infty}\frac{\left(1+\frac3x\right)^{1/3}-\left(1-\frac2x\right)^{1/2}}{1/x}\\ &=\lim_{x\to\infty}\frac{\frac13\left(1+\frac3x\right)^{-2/3}\left(-\frac3{x^2}\right)-\frac12\left(1-\frac2x\right)^{-1/2}\left(\frac2{x^2}\right)}{-1/x^2}\\ &=\lim_{x\to\infty}\left(\left(1+\frac3x\right)^{-2/3}+\left(1-\frac2x\right)^{-1/2}\right)\\ &=2 \end{align*}$$

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Our expression is equal to $$(\sqrt[3]{x^3+3x^2}-x)+(x-\sqrt{x^2-2x}.$$ Find the limits separately. The case of $x-\sqrt{x^2-2x}$ is the more standard one. one. Multiply by $\frac{x+\sqrt{x^2-2x}}{x+\sqrt{x^2+2x}}$. It will turn out that the linit is $1$.

The case of $\sqrt[3]{x^3+3x^2}-x$ is similar. Let $u=\sqrt[3]{x^3-2x}$ and $v=x$. Multiply by $\frac{u^2+uv+v^2}{u^2+uv+v^2}$. This exploits the fact that $(u-v)(u^2+uv+v^2)=u^3-v^3$. It will turn out that the limit is $1$.

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