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Problem: Evaluate the limit as $n$ approaches infinity of $\sqrt[3]{n^3+6n^2+36n+216}-\sqrt[3]{n^3+3n^2+9n+27}.$

If you approximate $n^3+6n^2+36n+216$ as $(n+2)^3$ and $n^3+3n^2+9n+27$ as $(n+1)^3,$ this limit evaluates to $1,$ which is what WolframAlpha gives.

However, if you approximate both as $n^3,$ the limit becomes $0.$

How do you know which approximation to use?

ada
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  • Where are you getting $(n+2)^3$ and $(n+1)^3$? By that logic I could also approximate them as $(n+101)^3$ and $(n+100)^3$ and still get the "right answer" but definitely not for any real mathematical reason. – Ninad Munshi Jun 12 '23 at 15:51
  • Well, $(n+2)^3 = n^3+6n^2+12n+8$ and $(n+1)^3=n^3+3n^2+3n+1,$ so these are closer to the expressions under the square root than $n^3.$ I guess my question is how do you actually evaluate this limit? – ada Jun 12 '23 at 15:55
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    Hint: Both roots are cube roots. This means we can use the difference of cubes formula: $$(a-b)(a^2+ab+b^2)=a^3-b^3$$ What would $a$ and $b$ have to be in this case? – Ninad Munshi Jun 12 '23 at 15:57
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    The way to use the hint of @NinadMunshi is to multiply the difference $a-b$ by $1$ in a convenient form. The hint suggests an expression for $1$ that is likely to be convenient. – Robert Shore Jun 12 '23 at 16:01
  • Setting $a=\sqrt[3]{n^3+6n^2+12n+8}$ and $b=\sqrt[3]{n^3+3n^2+9n+27},$ so then the answer is the limit as n approaches infinity of $\frac{a^3-b^3}{a^2+ab+b^2},$ which is $1.$ Is that correct? Follow-up question: How come approximating the limit as $\sqrt[3]{n^3}-\sqrt[3]{n^3}=0$ does not work? – ada Jun 12 '23 at 16:29
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    This (applied to $n=3,a_n=b_n=1,a_{n-1}=6,b_{n-1}=3$) answers your question: How to compute the limit of this difference of the n'th root of polynimials?. In that post, your variable supposedly integer $n\to+\infty$ is generalized to a real number $x\to+\infty.$ – Anne Bauval Jun 12 '23 at 19:02
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    Yep that's right! The answers below also give good insights to the problem solving here. To answer your follow up - short answer: $\infty - \infty$ is am indeterminate form. Long answer: It's generally a rule of thumb that when the dominant term cancels, the behavior of the next dominant term takes over (the next highest power in the case of $\to\infty$ or the next lowest power in the case of $\to0$). This is the idea behind L'Hopital's rule because taking derivatives gets you to the next dominant term. – Ninad Munshi Jun 12 '23 at 21:17
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    But just like how we sometimes need to repeat our application of L'Hopital's rule, sometimes limits are arranged in such a way that multiple levels of dominant terms cancel out. It is not always possible to readily see how far one needs to take an approximation for it to be valid, but generally you look for when "terms are equal". For example with your cube roots, the next dominant terms of $n^2$ had different coefficients so they were unlikely to cancel. This is just a heuristic though and does not always work as a naive application. – Ninad Munshi Jun 12 '23 at 21:20
  • That makes a lot of sense, thank you so much! – ada Jun 13 '23 at 01:41

2 Answers2

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If you are going to "approximate/ignore" parts of your formulas then you must keep track of the effect of these "ignored" parts on the calculation.

In your case, for $\sqrt[3]{n^3+6n^2+36n+216}$ you have the Laurent series

$$ \sqrt[3]{n^3+6n^2+36n+216} = n + 2 + \frac8n + \frac{112}{3 n^2} - \frac{736}{3 n^3} + O\left(\frac1{n^4}\right), $$

while for $\sqrt[3]{n^3+3n^2+9n+27}$ you have

$$ \sqrt[3]{n^3+3n^2+9n+27} = n + 1 + \frac2n + \frac{14}{3 n^2} - \frac{46}{3 n^3} + O\left(\frac1{n^4}\right). $$

You might not want to work out these series to so many terms. (I had Wolfram Alpha do it for me.) But you can relatively easily work out that

$$ \sqrt[3]{n^3+6n^2+36n+216} = n + 2 + O\left(\frac1n\right), $$ and $$ \sqrt[3]{n^3+3n^2+9n+27} = n + 1 + O\left(\frac1n\right). $$

You have already done this except for the $O\left(\frac1n\right)$ part.

If you take the difference of two expressions that contain $O\left(\frac1n\right)$ terms, you get an $O\left(\frac1n\right)$ term in the result:

$$ \left(n + 2 + O\left(\frac1n\right)\right) - \left(n + 1 + O\left(\frac1n\right)\right) = 1 + O\left(\frac1n\right). $$

This tells you that

$$ \sqrt[3]{n^3+6n^2+36n+216}-\sqrt[3]{n^3+3n^2+9n+27} = 1 + O\left(\frac1n\right), $$

which is easy to evaluate in the limit as $n\to\infty$ because $\frac1n\to 0.$

On the other hand, you could make an even coarser approximation,

$$ \sqrt[3]{n^3+6n^2+36n+216} = n + O(1) $$ and $$ \sqrt[3]{n^3+3n^2+9n+27} = n + O(1). $$

This is what you are doing when you "approximate both as $n^3$".

Now if you take the difference of two $O(1)$ terms, the result is another $O(1)$ term, so this method gives you the result

$$ \sqrt[3]{n^3+6n^2+36n+216}-\sqrt[3]{n^3+3n^2+9n+27} = (n + O(1)) - (n + O(1)) = O(1), $$

which is generally not zero. It is something that might converge to a constant as $n\to\infty$, but this method doesn't tell you what that constant is.

In summary, the reason you know you can "ignore" terms of the approximation after the constant term (for example, $\sqrt[3]{n^3+6n^2+36n+216} \approx n + 2$) is because you know all of those terms -- and the differences of those terms -- will go to zero as $n \to \infty.$ The reason you cannot "ignore" the $2$ in this approximation is because the difference of constant terms does not go to zero as $n \to \infty.$

As for how you might know to write $\sqrt[3]{n^3+6n^2+36n+216} = n + 2 + O\left(\frac1n\right)$, first consider how you know that $n + 2$ is a better approximation for $\sqrt[3]{n^3+6n^2+36n+216}$ than $n + 1$ is. Rigorously speaking, there are probably multiple ways to do this, but one way is to use the multinomial theorem (a generalization of the binomial theorem for powers of sums of more than two terms) to the expression

$$ (n^3+6n^2+36n+216)^{1/3}, $$

which (like the expansion of a fractional power of a binomial) produces an infinite series. You don't actually need to compute more than the first couple of terms of that series for this problem; all you need to know is that the first term is $n$, the second term is $2$, the third term is something times $\frac1n$, and the remaining terms are various constants times higher powers of $\frac1n$.

Less rigorously, but perhaps more intuitively, you can try to figure out what you would add to $n + 2$ so that when you cube your sum you get the first three terms of $n^3+6n^2+36n+216$ exactly instead of just the first two terms.

David K
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  • Thank you! that is helpful. But how would I know that without any tools like Wolfram, just from looking? – ada Jun 12 '23 at 16:47
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$$\sqrt[3]{n^3+6n^2+36n+216}-\sqrt[3]{n^3+3n^2+9n+27}=\dfrac{n^3+6n^2+36n+216-n^3-3n^2-9n-27}{(n^3+6n^2+36n+216)^{2/3}+(n^3+3n^2+9n+27)^{2/3}+(n^3+6n^2+36n+216)^{1/3}(n^3+3n^2+9n+27)^{1/3}}\sim \frac{3n^2}{3n^2}=1$$

Harish Chandra Rajpoot
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Guillermo García Sáez
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