Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$

Question

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

Answer 1

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.

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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.

Answer 2

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 $$\sqrt[3]{x (x+1) (x+2)}\sim 1 + x;\,x\to\infty$$

Answer 3

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.

Answer 4

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

$$=x \left(1+\frac1x+O\left(\frac1{x^2}\right)-2\left(1+\frac1x\right)+1+\frac1x+O\left(\frac1{x^2}\right)\right)=O\left(\frac1{x}\right) \to 0$$