# Evaluate $\lim\limits_{x\to\infty}x\!\left[2x\!-\!\left(x^3\!+\!x^2\!+\!x\right)^{\!\frac13}\!\!-\!\left(x^3\!-\!x^2\!+\!x\right)^{\!\frac13}\right]$ [closed]

Evaluate $$\lim\limits_{x\to \infty}x\left[2x-\left(x^3+x^2+x\right)^{\frac{1}{3}}-\left(x^3-x^2+x\right)^{\frac{1}{3}}\right]$$

My Approach:

Formula I used $$(1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^2+.....\infty$$ where $$x\in(-1,1)$$

$$\lim\limits_{x\to \infty}x\left[2x-x\left(1+\left(\frac{1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}-x\left(1+\left(\frac{-1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}\right]$$

$$\implies\lim\limits_{x\to \infty}x\left[2x-x\left(1+\left(\frac{1}{3x}+\frac{1}{3x^2}\right)\right)-x\left(1+\left(\frac{-1}{3x}+\frac{1}{3x^2}\right)\right)\right]$$

$$\implies\lim\limits_{x\to\infty}x\left(-\frac{2}{3x}\right)=-\frac{2}{3}$$

But given answer is $$\frac{2}{9}$$. I am attaching given solution below. Also I am attaching my two more solutions in Image formart.

• Note that Wolfram gives the answer as -4/9. You need to be more careful with tracking down the terms, and using enough expansions. Note that in your handwritten version, you didn't distribute the $x$ throughout when expanding $x(1 + ( \pm 1/x + 1/x^2) ) ^ {1/3}$. So you should get $- 1/3 + 1/9 - 1/3 + 1/9 = -4/9$ , which agrees with WA. Commented May 18 at 6:47
• I agree that $I_2 = 2025$, and thus conclude that none of the 4 options are correct. Commented May 18 at 6:51
• @CalvinLin Also can you tell me what is wrong in method used in last solution? Commented May 18 at 10:44
• I don't understand how you went from limits $[(1+t + t^2)^{1/3} - 1^{1/3} ] / t^2 = 1/3 \times ( t + t^2 ) / t^2$. You seem to be assuming that limits $[ 1 + f(x) ]^\alpha - 1 ^ \alpha = \alpha f(x)$, which doesn't hold for quadratics and higher. $\quad$ You're not properly accounting for the higher order terms, esp since you're dividing out by $t^2$. Commented May 18 at 17:02

The formula you used is correct, however you made a couple of mistakes when computing the expansion at infinity, in fact you have: $$\left(1+\frac 1x+\frac 1{x^2}\right)^{1/3}\sim 1+\frac 1{3x}+\frac 2{9x^2} \\ \left(1-\frac 1x+\frac 1{x^2}\right)^{1/3}\sim 1-\frac 1{3x}+\frac 2{9x^2}$$ Hence \begin{align} &\lim_{x\to \infty} x\left[2x-x\left(1+\left(\frac{1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}-x\left(1+\left(\frac{-1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}\right]= \\ &\lim_{x\to \infty} x\left[2x-x\left(1+\frac 1{3x}+\frac 2{9x^2}\right)-x\left(1-\frac 1{3x}+\frac 2{9x^2}\right)^{\frac{1}{3}}\right]= \\ &\lim_{x\to \infty} x\left(2x-x-\frac13-\frac 2{9x} -x+\frac13-\frac 2{9x} \right)= -\frac49 \end{align}

To be very safe, a step by step approach could be followed to do this. First we need this expansion: $$\sqrt[{\large 3}]{1+y} = 1 +\frac{y}{3} -\frac{y^2}9 +\frac{5y^3}{81}+ \dots, \ \ \ \text{if} \ y\rightarrow 0,$$ where we actually have one term more than we expect to be necessary. Then this is used to expand these two cases: \begin{align} \sqrt[{\large 3}]{1+\frac1 x +\frac 1 {x^2}} &= 1 +\frac{1}{3x} +\frac{2}{9x^2} -\frac{13}{81x^3}+ \dots, \\[5pt] \sqrt[{\large 3}]{1-\frac1 x +\frac 1 {x^2}} &= 1 -\frac{1}{3x} +\frac{2}{9x^2} +\frac{13}{81x^3}+ \dots, \ \ \ \text{if} \ x\rightarrow \infty \end{align} where we again keep one term more than needed. Then of course the result follows quickly: \begin{align} &x^2\left(2-\sqrt[{\large 3}]{1+\frac1 x +\frac 1 {x^2}}- \sqrt[{\large 3}]{1-\frac1 x +\frac 1 {x^2}}\right) = \\[6pt] x^2\Big(&-\frac{4}{9x^2} +O\big(\frac1{x^4}\big)\Big) =-\frac{4}{9} +O\big(\frac1{x^2}\big), \ \ \ \text{if} \ x\rightarrow \infty \end{align} which proves that the limit we want is $$-\frac{4}{9}$$. But to find those expansions and check how many terms we need to keep is a lot of work, unless you leave it to a symbolic calculator.

So is there a fast way to see the answer immediately off the top of your head? This might be possible if you first observe that we need: \begin{align} \lim_{x\rightarrow\infty}\ &x^2\Big(2-f\big(1+\frac1 x +\frac 1 {x^2}\big)- f\big(1-\frac1 x +\frac 1 {x^2}\big)\Big), \end{align} with $$f(y)=\sqrt[{\large 3}]{y}$$, which I think most people will see, and then have ready knowledge of relations like: \begin{align} f\big(1+ \delta_1 \big) &= f(1) + f'(1)\ \delta_1 \\[6pt] f\big(1+ \delta_2 \big) + f\big(1- \delta_2 \big) &= 2 f(1) + f''(1)\ \delta_2^2, \end{align} which probably also is not asking too much (although the missing factor $$2$$ in the second one is already treacherous). But then you finally need the experience to see that odd and even terms do not interfere and you can handle two $$\delta$$'s like this: $$f\big(1+ \delta_2 +\delta_1\big) + f\big(1- \delta_2+\delta_1 \big) = 2 f(1) + 2f'(0)\ \delta_1+ f''(0)\ \delta_2^2,$$ and in addition you have to see that in this case the function $$f$$, i.e. the cube root, has $$f'(1)=\frac13$$ and $$f''(1)=-\frac29$$. That would give you $$-\frac23+\frac29$$, so you get the correct answer.

I think that given the minus signs and factors $$2$$ involved here, it would be a bit of a gamble, but that the more experienced mathematicians (at least those who like this game) would be able to do it!

We need to consider also the second order term in the expansion giving an extra term $$\frac2{9x}$$ before simplifying by $$x$$ term, to obtain

$$\ldots \implies \lim_{x\to\infty}x\left(-\frac{2}{3x}+\frac2{9x}\right)=-\frac{4}{9}$$

The issue is that when we use series expansion in limits the remainder term should always be indicated, in this case we are keeping second order terms and then we can’t truncate the expansion at the first order.

For example, expanding the first term and using little $$o$$ notation, we should indicate

$$-x\left(1+\left(\frac{1}{x}+\frac{1}{x^2}\right)\right)^{\frac{1}{3}}=-x\left(1+\frac13\left(\frac{1}{x}+\frac{1}{x^2}\right)-\frac29\left(\frac{1}{x}+\frac{1}{x^2}\right)^2+o\left(\frac1{x^2}\right)\right)=$$

$$\require{cancel} =-x\left(1+\frac1{3x}+\frac{1}{3x^2}-\frac{2}{9x^2}\color{red}{\cancel{-\frac{4}{9x^3}}\cancel{-\frac{2}{9x^4}}}+o\left(\frac1{x^2}\right)\right)$$