<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] first of all, notice that the substitution $x+1 = t, t \to \infty$ simplifies things quite a bit:
$$L = \lim_{t \to \infty} \left( (t(t^2-1))^{1/3} + ((t-1)^{t-1}\cdot t^t \cdot (t+1)^{t+1})^{1/3t} - 2t\right)$$
Factor out $t$ from the expression in the bracket and do some manipulations to get
$$\begin{align}
L &= \lim_{t \to \infty}\ t\left( \left( 1 - \frac{1}{t^2}\right)^\frac13 + \left( \frac{t+1}{t-1}\right)^\frac{1}{3t} \cdot \left(1 - \frac{1}{t^2}\right)^\frac13 - 2\right) \\ 
&= \lim_{t \to \infty}\ t\left( \left( 1 + \left( \frac{t+1}{t-1}\right)^\frac{1}{3t} \right)\left( 1 - \frac{1}{t^2}\right)^\frac13 - 2\right)
\end{align}$$
The term in the first bracket tends to 2 as $t$ tends to infinity, so we're finally getting somewhere. All we have to do is prove that the order of the infinitesimal within the bracket is $\frac{1}{t^2}$ or lesser. To do this, notice that
$$\left( \frac{t+1}{t-1}\right)^\frac{1}{3t} = \left( 1 + \frac{2}{t-1}\right)^\frac{1}{3t} \approx 1 + \frac{2}{3t(t-1)}$$
(By using binomial expansion). We can also assume that the second term tends to 1, i.e.
$$\left( 1 - \frac{1}{t^2}\right)^\frac13 \approx 1$$
(I'll justify this in a minute, but let's carry on for now)
once we have done these simplifications, the limit becomes
$$\begin{align}
L &= \lim_{t \to \infty}\ t\left( 2 + \frac{2}{3t(t-1)} - 2\right) \\
&= \lim_{t \to \infty}\ \frac{2t}{3t(t+1)} \\
&= \boxed{0}
\end{align}$$
Hence proved.

For justifying the approximation(s) I did, let's look at the expression and try to simplify via binomial theorem to the first order:
$$\begin{align}
f(t) &= \left( 2 + \frac{2}{3t(t-1)}\right)\left( 1 - \frac{1}{t^2}\right)^\frac13 \\
&= \left( 2 + \frac{2}{3t(t-1)}\right)\left( 1 - \frac{1}{3t^2}\right)\\
&= 2 - \frac{2}{3t^2} + \frac{2}{3t(t-1)} - \frac{2}{9t^3(t-1)} \\
&= 2 + \frac{2}{3t^2(t-1)} - \frac{2}{9t^3(t-1)}
\end{align}$$
You notice that we get a different expression here, one that has a $t^2$ instead of a $t$ in the denominator. This doesn't matter as the power of $t$ in the denominator should be more than 1 to obtain the result as 0. While this expansion gives a more accurate answer, the approximation that I did in the answer is justified.
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