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?
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:
$$\begin{align}\lim_{x\to +\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})
&amp;=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-\sqrt{1\ -\frac{2}{x}}\right)
\\&amp;=\lim_{x\to +\infty}x\left(\sqrt[3]{1+\frac{3}{x}}-1\right)-\lim_{x\to \infty}x\left(\sqrt[2]{1 +\frac{-2}{x}}-1\right)\\&amp;=\frac{3}{3}-\frac{-2}{2}=1+1=2.
\end{align}$$

P.S. Note that on the other side,
$$\lim_{x\to -\infty} (\sqrt[3]{x^{3}+3x^{2}}-\sqrt{x^{2}-2x})=-\infty$$
From OP's work
$$L=\lim_{x \rightarrow \infty} x \left[ \left( 1+\frac{3}{x} \right)^{1/3}-\left( 1-\frac{2}{x} \right)^{1/2}\right]= \lim_{x \rightarrow \infty} x \left[ \left( 1+\frac{3}{3x} \right)-\left( 1-\frac{2}{2x} \right)\right]= \lim_{x \rightarrow \infty}x \frac{2}{x}=2. $$