Find $\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$ without L'Hopital or Taylor series. 
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$

My try is as follows:
$$\large \lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=$$$$ \lim_{x\to 
∞}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$$$=\lim_{x\to ∞}x\lim_{x\to 
∞}\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)$$
which is $∞×0$ , but clearly this zero is not exactly zero. I was thinking about generalized binomial 
theorem, but seems it will make the limit difficult, so how this kind of limits can be solved without using Taylor series or L'Hopital's rule?
 A: We first note that for any positive integer $n$ and any real $a$,
$$\lim_{x\to \infty}x\left(\sqrt[n]{1+\frac{a}{x}}-1\right)=
\lim_{s\to 1}a\frac{s-1}{s^n-1}=\lim_{s\to 1}\frac{a}{s^{n-1}+s^{n-2}+\dots +s +1}=\frac{a}{n}$$
where $s=\sqrt[n]{1+a/x}$ and therefore $a/x=s^n-1$, and $x=a/(s^n-1)$. 
Hence, from your work, we split the limit in two:
$$\begin{align}\lim_{x\to +\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})
&=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)
\\&=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-1\right)-\lim_{x\to \infty}x\left(\sqrt[2]{1 +\frac{-2}{x}}-1\right)\\&=\frac{3}{3}-\frac{-2}{2}=1+1=2.
\end{align}$$
P.S. Note that on the other side,
$$\lim_{x\to -\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=-\infty$$
A: Using high-school math:$$\lim_{x\to ∞} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=\lim_{x\to ∞} \frac{\sqrt[3]{(x^{3}+3x^{2})^2}-x^{2}+2x}{\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x}}=$$ $$ \lim_{x\to ∞} \frac{(x^{3}+3x^{2})^2-(x^{2}-2x)^3}{(\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x})(\sqrt[3]{(x^{3}+3x^{2})^4}+\sqrt[3]{(x^{3}+3x^{2})^2}(x^{2}-2x)+(x^2-2x)^2)}=$$ $$\lim_{x\to ∞} \frac{12x^5-3x^4+8x^3}{(\sqrt[3]{x^{3}+3x^{2}}+\sqrt{x^{2}-2x})(\sqrt[3]{(x^{3}+3x^{2})^4}+\sqrt[3]{(x^{3}+3x^{2})^2}(x^{2}-2x)+(x^2-2x)^2)}=$$ $$[\text{leaving the highest power}]=\lim_{x\to ∞} \frac{12x^5}{(x+x)(x^4+x^4+x^4)}=2$$
A: A quite elementary way is just using the two binomial formulas $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$ as follows:
\begin{eqnarray*}\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x}
& = & (\sqrt[3]{x^{3}+3x^{2}} - x) + (x -\sqrt{x^{2}-2x})\\
& = & \frac{3x^2}{\sqrt[3]{(x^{3}+3x^{2})^2} + x\sqrt[3]{x^{3}+3x^{2}} + x^2} + \frac{2x}{x+\sqrt{x^{2}-2x}}\\
& = & \frac{3}{\sqrt[3]{(1+\frac{3}{x})^2} + \sqrt[3]{1+\frac{3}{x}} + 1} + \frac{2}{1+\sqrt{1-\frac{2}{x}}} \\
& \stackrel{x\to \infty}{\longrightarrow} & 1+1 = 2
\end{eqnarray*}
A: By binomial approximation


*

*$\sqrt[3]{x^{3}+3x^{2}}=x\sqrt[3]{1+3/x}\approx x\left(1+\frac1x\right)=x+1$

*$\sqrt{x^{2}-2x}=x\sqrt{1-2/x}\approx x\left(1-\frac1x\right)=x-1$
therefore
$$\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x}\approx 2$$
A: Hint:
Set $1/n=h$
$$x^3+3x^2=\dfrac{1+3h}{h^3}, x^2-2x=\dfrac{1-2h}{h^2}$$
$$\large \lim_{x\to\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})$$
$$=\lim_{h\to0^+}\dfrac{\sqrt[3]{1+3h}-\sqrt{1-2h}}h$$
$$=\lim_{h\to0^+}\dfrac{\sqrt[3]{1+3h}-1}h-\lim_{h\to0^+}\dfrac{\sqrt{1-2h}-1}h$$
Set $\sqrt[3]{1+3h}-1=p\implies1+3h=(1+p)^3, h=p+p^2+\dfrac{p^3}3$
and $\sqrt{1-2h}-1=q\implies1-2h=(1+q)^2,-h=q+\dfrac{q^2}2$
A: From OP's work
$$L=\lim_{x \rightarrow \infty} x \left[ \left( 1+\frac{3}{x} \right)^{1/3}-\left( 1-\frac{2}{x} \right)^{1/2}\right]= \lim_{x \rightarrow \infty} x \left[ \left( 1+\frac{3}{3x} \right)-\left( 1-\frac{2}{2x} \right)\right]= \lim_{x \rightarrow \infty}x \frac{2}{x}=2. $$
