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? 
 A: 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:
$$\lim_{x\to \infty}{\frac{(x^3+6x^2+11x+6)-x^3}{\sqrt[3]{(x+1)(x+2)(x+3)}^2+x\sqrt[3]{(x+1)(x+2)(x+3)}+x^2}}\frac{1/x^2}{1/x^2}=$$
$$\lim_{x\to \infty}{\frac{6x^2+11x+6}{\sqrt[3]{(x+1)(x+2)(x+3)}^2+x\sqrt[3]{(x+1)(x+2)(x+3)}+x^2}}\frac{1/x^2}{1/x^2}=$$
$$\lim_{x\to \infty}{\frac{6+11/x+6/x^2}{\sqrt[3]{(1+1/x)(1+2/x)(1+3/x)}^2+(1)\sqrt[3]{(1+1/x)(1+2/x)(1+3/x)}+1}}=$$
$$\frac{6}{\sqrt[3]{(1)(1)(1)}^2+\sqrt[3]{(1)(1)(1)}+1}=\frac{6}{3}=2$$
A: Let $1/x=h$
$$\lim_{h\to0^+}\dfrac{\sqrt[3]{(1+h)(1+2h)(1+3h)}-1}h$$
$$=\lim_{h\to0^+}\dfrac{(1+h)(1+2h)(1+3h)-1}h\cdot\lim_{h\to0^+}\dfrac1{1+\sqrt[3]{(1+h)(1+2h)(1+3h)}+\sqrt[3]{(1+h)^2(1+2h)^2(1+3h)^2}}$$
A: $\sqrt[3]{(x+1)(x+2)(x+3)}=x\sqrt[3]{(1+1/x)(1+2/x)(1+3/x)}
=x(1+6/x+O(x^{-2}))^{1/3}=x(1+2/x+O(x^{-2}))$.
A: Using general Binomial expansion, $((x+1)(x+2)(x+3))^\frac{1}{3}-x = ((x+2)^3-(x+2))^\frac{1}{3}=(x+2)-\frac{1}{3}(x+2)^{-2}(x+2)-\frac{1}{9}(x+2)^{-5}(x+2)^3+..-x=2-P(x+2)$, 
where the power of the polynomial $P(x+2)$ is negative. Letting $x \to \infty$, we get the limit 2.
A: $y = x+2$:
$$L = \lim _{x\to \infty} (\sqrt[3]{(x+1)(x+2)(x+3)}-x) = 2+ \lim _{y\to \infty} (\sqrt[3]{y^3-y}-y) = 2+ \lim _{y\to \infty} \frac{\sqrt[3]{1-\frac{1}{y^2}}-1}{\frac{1}{y}}$$
$y = \frac{1}{h}$:
$$L = 2+ \lim _{h\to 0^+}\frac{\sqrt[3]{1-h^2}-1}{h} \stackrel{L'Hopital}{=}2+ \lim _{h\to 0^+}\frac{\frac{-2h}{\sqrt[3]{(1-h^2)^2}}}{1} = 2+0 = 2$$
