<s>[INST] Evaluating $\lim _{x\to \infty} (\sqrt[3]{(x+1)(x+2)(x+3)}-x)$ 
Evaluate:  $\lim _{x\to \infty}( \sqrt[3]{(x+1)(x+2)(x+3)}-x)$

I set $y= x-3$ for simplification and then tried to solve $y\to \infty$. 
I tried to use the tayor expansion of the cubic function f(y) in the cuberoot but that didn't help. 
How do I approach this problem then? 
 [/INST] The limit is based upon the difference of the cube identity:
$$\lim_{x\to \infty}{\Big(\sqrt[3]{(x+1)(x+2)(x+3)}-x\Big)}=$$
$$\lim_{x\to \infty}{\Big(\sqrt[3]{(x+1)(x+2)(x+3)}-x\Big)\frac{\sqrt[3]{(x+1)(x+2)(x+3)}^2+x\sqrt[3]{(x+1)(x+2)(x+3)}+x^2}{\sqrt[3]{(x+1)(x+2)(x+3)}^2+x\sqrt[3]{(x+1)(x+2)(x+3)}+x^2}}$$
$$=\lim_{x\to \infty}{\frac{(x+1)(x+2)(x+3)-x^3}{\sqrt[3]{(x+1)(x+2)(x+3)}^2+x\sqrt[3]{(x+1)(x+2)(x+3)}+x^2}}=$$
Now divide both the numerator and denominator by the greatest power of the denominator:
$$\lim_{x\to \infty}{\frac{(x^3+6x^2+11x+6)-x^3}{\sqrt[3]{(x+1)(x+2)(x+3)}^2+x\sqrt[3]{(x+1)(x+2)(x+3)}+x^2}}\frac{1/x^2}{1/x^2}=$$
$$\lim_{x\to \infty}{\frac{6x^2+11x+6}{\sqrt[3]{(x+1)(x+2)(x+3)}^2+x\sqrt[3]{(x+1)(x+2)(x+3)}+x^2}}\frac{1/x^2}{1/x^2}=$$
$$\lim_{x\to \infty}{\frac{6+11/x+6/x^2}{\sqrt[3]{(1+1/x)(1+2/x)(1+3/x)}^2+(1)\sqrt[3]{(1+1/x)(1+2/x)(1+3/x)}+1}}=$$
$$\frac{6}{\sqrt[3]{(1)(1)(1)}^2+\sqrt[3]{(1)(1)(1)}+1}=\frac{6}{3}=2$$
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