<s>[INST] Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\infty}\frac{(x(x+1)(x+2))^{\frac{1}{3}}-2(x+1)}{-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}}=1$$
Using the Hospital rule but it doesn't help here .Moreover I have tried power series without success .
Any helps is welcome .
Thanks in advance
 [/INST] Expanding at $\infty$ with Taylor formula, we get
$$x^{\frac{x}{3 x+3}} (x+1)^{\frac{x+1}{3 x+3}} (x+2)^{\frac{x+2}{3 x+3}}\sim 1+x;\;x\to\infty$$
and
$$\sqrt[3]{x (x+1) (x+2)}\sim 1+x;\;x\to\infty$$
Therefore as $x\to\infty$ limit becomes
$$1+x-(2(x+1)-(1+x))=0$$
$$.....................................$$
To expand at $\infty$ we plug $1/x=y$ and expand at $y=0$
For instance
$$\sqrt[3]{x (x+1) (x+2)}=\frac{1}{y}\sqrt[3]{(y+1) (2 y+1)}$$
Expand at $y=0$
$$\sqrt[3]{(y+1) (2 y+1)}= 1+y+O\left(y^2\right)$$
plug again $y=1/x$ to get
$$\sqrt[3]{x (x+1) (x+2)}=x\left(1+\frac{1}{x}\right)+O\left(1/x^2\right)$$
therefore
$$\sqrt[3]{x (x+1) (x+2)}\sim 1 + x;\,x\to\infty$$
</s>