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
 A: 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.
A: It suffices to note that as $x\to +\infty$,
$$\left(1+\frac{a}{x}\right)^{m\frac{x+b}{x+c}}=
\exp\left(m\frac{1+b/x}{1+c/x}\log\left(1+\frac{a}{x}\right)\right)=1+\frac{ma}{x}+o(1/x).$$
Then,
\begin{align*}(x(x+1)(x+2))^{\frac{1}{3}}&=x\left(1+\frac{1}{x}\right)^{\frac{1}{3}}\left(1+\frac{2}{x}\right)^{\frac{1}{3}}\\
&=x\left(1+\frac{1}{3x}+\frac{2}{3x}+o(1/x)\right)=x+1+o(1)
\end{align*}
and
\begin{align*}\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}&=
x\left(1+\frac{1}{x}\right)^{\frac{x+1}{3x+3}}\left(1+\frac{2}{x}\right)^{\frac{x+2}{3x+3}}\\
&=x\left(1+\frac{1}{3x}+\frac{2}{3x}+o(1/x)\right)=x+1+o(1).
\end{align*}
Then
$$(x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\\=
x+1+o(1)-\Big(2(x+1)-\Big(x+1+o(1)\Big)=o(1)
$$
and the given limit easily follows.
A: 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
$$\sqrt[3]{x (x+1) (x+2)}\sim 1 + x;\,x\to\infty$$
A: We have that
$$\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=$$
$$=x\left(\left(1+\frac1x\right)^{\frac{1}{3}}\left(1+\frac2x\right)^{\frac{1}{3}}-\left(2\left(1+\frac1x\right)-\left(\left(1+\frac1x\right)^{x+1}\left(1+\frac2x\right)^{x+2}\right)^{\frac{1}{3x+3}}\right)\right)=$$
and since by binomial expansion we have

*

*$\left(1+\frac1x\right)^{\frac{1}{3}}\left(1+\frac2x\right)^{\frac{1}{3}}=\left(1+\frac1{3x}+O\left(\frac1{x^2}\right)\right)\left(1+\frac2{3x}+O\left(\frac1{x^2}\right)\right)=1+\frac1x+O\left(\frac1{x^2}\right)$

*$\left(\left(1+\frac1x\right)^{x+1}\left(1+\frac2x\right)^{x+2}\right)^{\frac{1}{3x+3}}=\left(1+\frac1x\right)^{\frac13}\left(1+\frac2x\right)^{\frac13}\left(1+\frac2x\right)^{\frac1{3x+3}}=1+\frac1x+O\left(\frac1{x^2}\right)$
then
$$=x \left(1+\frac1x+O\left(\frac1{x^2}\right)-2\left(1+\frac1x\right)+1+\frac1x+O\left(\frac1{x^2}\right)\right)=O\left(\frac1{x}\right) \to 0$$
