Easy Limit

Well this problem becomes very easy if we use (1+x)^ n = 1+nx with x approaches zero

Take x^3 and x^2 common respectively from the two radicals and use above identity to get

x(1+1/x)-x(1-1/x) = 2

First I am going to expand this expression to lose the square root, doing that by using the identity $(a-b)(a+b)=a^2-b^2$ :

$\displaystyle\lim_{x\to\infty} \left( \sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x} \right) =\displaystyle\lim_{x\to\infty} \left( \{\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\}\color{#3D99F6}{\frac{\sqrt[3]{x^3+3x^2}+\sqrt{x^2-2x}}{ \sqrt[3]{x^3+3x^2}+\sqrt{x^2-2x}}} \right)$

After multiplying, we get the following expression:

$\displaystyle\lim_{x\to\infty} \left(\frac{\sqrt[3]{(x^3+3x^2)^2}-x^2-2x}{ \sqrt[3]{x^3+3x^2}+\sqrt{x^2-2x}} \right)$

Now I am going to expand again, this time to lose the 3rd root, doing that by using the identity $(a-b)(a^2+ab+b^2)=a^3-b^3$ :

$\displaystyle\lim_{x\to\infty} \left( \frac{\sqrt[3]{(x^3+3x^2)^2}-x^2-2x}{ \sqrt[3]{x^3+3x^2}+\sqrt{x^2-2x}}\color{#3D99F6}{\frac{\sqrt[3]{(x^3+3x^2)^4}+(x^2-2x)\sqrt[3]{(x^3+3x^2)^2}+(x^2-2x)^2}{\sqrt[3]{(x^3+3x^2)^4}+(x^2-2x)\sqrt[3]{(x^3+3x^2)^2}+(x^2-2x)^2} }\right)$

After expanding we get:

$\displaystyle\lim_{x\to\infty} \left(\frac{(x^3+3x^2)^2-(x^2-2x)^3}{(\sqrt[3]{(x^3+3x^2)^4}+(x^2-2x)\sqrt[3]{(x^3+3x^2)^2}+(x^2-2x)^2)( \sqrt[3]{x^3+3x^2}+\sqrt{x^2-2x})} \right)$

Now by expanding the numerator of this fraction, and dividing (and multiplying) the denominator with $x^5$ , we get

$\displaystyle\lim_{x\to\infty} \left(\frac{12x^5-3x^4+8x^3}{\color{#3D99F6}{x^5}(\sqrt[3]{(1+\frac{3}{x})^4}+(1-\frac{2}{x})\sqrt[3]{(1+\frac{3}{x})^2}+(1-\frac{2}{x})^2)( \sqrt[3]{1+\frac{3}{x}}+\sqrt{1-\frac{2}{x}})} \right)$

Now by dividing(and multiplying) the nominator with $x^5$ we get:

$\displaystyle\lim_{x\to\infty} \left(\frac{\color{#D61F06}{x^5}(12-\frac{3}{x}+\frac{8}{x^2})}{\color{#D61F06}{x^5}(\sqrt[3]{(1+\frac{3}{x})^4}+(1-\frac{2}{x})\sqrt[3]{(1+\frac{3}{x})^2}+(1-\frac{2}{x})^2)( \sqrt[3]{1+\frac{3}{x}}+\sqrt{1-\frac{2}{x}})} \right)$

And finally, by using the limits $\lim_{x\to\infty} { \frac{1}{x}}=0$ and $\lim_{x\to\infty}\frac{1}{x^2}=0$ we get:

$\displaystyle\lim_{x\to\infty} \left(\frac{(12-\frac{3}{x}+\frac{8}{x^2})}{(\sqrt[3]{(1+\frac{3}{x})^4}+(1-\frac{2}{x})\sqrt[3]{(1+\frac{3}{x})^2}+(1-\frac{2}{x})^2)( \sqrt[3]{1+\frac{3}{x}}+\sqrt{1-\frac{2}{x}})} \right)=\frac{12}{(1+1+1)(1+1)}=\frac{12}{6}$

$\boxed{ \displaystyle\lim_{x\to\infty} \left( \sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x} \right) = 2}$

Use y=1/x and L´HOSPITAL rule.