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Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9}$.

My first thoughts are the that I can remove a factor of $2n$ from each root, and obtain

$$ \lim_{n\to\infty}2n\sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-2n\sqrt{1+\frac{1}{2n}+\frac{9}{4n^2}} $$

But factoring out $2n$ won't help evaluate the limit because of the indeterminate form... So then I thought of using L'Hop, but thinking about the chain rule there makes me think that its more effort than it's worth and there's a missing something I can't get to. Any suggestions?

Amaan M
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  • @L'HJospital riule is not a and idea, but first some algebraic manipulations are needed to set it up. – Mittens May 06 '21 at 06:21
  • @OliverDiaz would you care to elaborate? – C Squared May 06 '21 at 06:43
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    @CSquared: If you follow some of the manipulations that Paramanand did at the beginning, you see how you can get read of the annoying cubic and square roots. That gives rations of expressions on which you can try your luck with L'Hospital rule. – Mittens May 06 '21 at 06:47
  • Related: https://math.stackexchange.com/questions/30040/limits-how-to-evaluate-lim-limits-x-rightarrow-infty-sqrtnxna-n-1 – Hans Lundmark May 06 '21 at 07:13
  • @CSquared: the link by Hans Lundmark has what I had in mind. – Mittens May 06 '21 at 14:50

5 Answers5

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For small $x$, $(1+x)^a \approx 1+ax$. If you use this on both roots, the $1$s will cancel leaving you with a next term that is of order $\frac 1n$. That takes care of the $n$ out front and you get a constant.

Ross Millikan
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You were going on right track and shouldn't have stopped. There is also a typo when you take $2n$ out of square root. The terms in square root should be like $1+(1/4n)+\dots$.

Factoring out $2n$ allows us to rewrite the expression under limit as $$2n(\sqrt[3]{A}-\sqrt{B})$$ where $A, B$ are functions of $n$ tending to $1$ as $n\to\infty $. Now using this fact we add and subtract $1$ to get $$2n(\sqrt[3]{A}-1-(\sqrt{B}-1))$$ And this can be further rewritten as $$2n(A-1)\cdot\frac{\sqrt[3]{A}-1}{A-1}-2n(B-1)\cdot\frac {\sqrt{B} - 1}{B-1}$$ The fractions in above expression tend to $1/3,1/2$ respectively via the formula $$\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{1}$$ and observe that $$2n(A-1)=2n\left(\frac {1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}\right)\to 1$$ and similarly $$2n(B-1)\to \frac{1}{2}$$ The desired limit is thus $$1\cdot\frac{1}{3}-\frac{1}{2}\cdot\frac{1}{2}=\frac {1}{12}$$


Most usual limit problems related to algebraic functions don't need anything more than a little algebraic manipulation targeted at making use of the limit formula $(1)$. Advanced tools like Taylor and L'Hospital's Rule are best left for difficult limit problems.

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    I disagree with your last comment, but I respect your opinion, Taylor and L'Hospital rule are in fact basic tools one should encourage to use the even t to tie your shoes. – Mittens May 06 '21 at 06:21
  • @OliverDiaz: thanks for the feedback. My comment was directed to people learning limits for the first time. I don't really expect them to be aware of such tools as a part of their course. Often it happens that beginners are charmed by the use of such tools like L'Hospital's Rule without understanding them fully and often fall in various traps. If one is aware of how these tools work then it is better to use them to get the job done easily. – Paramanand Singh May 06 '21 at 06:25
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    Agreed, but the OP seems to be aware of the tools, he/she was not able to set it up correctly (for that some algebraic manipulations are needed) In any event, once you have the wheel you should not go back to carrying this by brute force. (the Egyptians and the Aztecs built great stuff without the wheel, but imagine if that had discover the wheel), their pyramids would have been more like geek palaces. – Mittens May 06 '21 at 06:30
  • @OliverDiaz: I like your analogy of Egyptians and Aztecs. It's very picturesque. +1 for that comment. – Paramanand Singh May 06 '21 at 06:32
  • @OliverDiaz, the use of the "A" and "B" expressions helped to see the algebraic manipulation that would really drive the limit home for me. I teach an AP Calc class and have some real talented students and like to challenge them at the end of the year with difficult limits and see how they respond. – Eleven-Eleven May 06 '21 at 16:09
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    @Eleven-Eleven: I have no quarrel with that. That is the power of abstraction. If your comment is with regards my friendly exchange with Paramanand, my point is that the earlier students understand the power of Taylor approximations and even L'Hospital's rule, and learn how to use them (algebraic manipulations will always be there) the better. For then they can move to very serious things: optimization problems, calculations in physics, etc. – Mittens May 06 '21 at 16:15
  • @OliverDiaz, yes, it was in regards to that. And I do understand myself the taylor and L'Hop... my student's know L'Hop but not Taylor. So I was looking for a way not involving L'Hop but was driven more by an algebraic approach which is why I was approaching it as such. Although, I can probably teach basic series (certainly taylor, i usually explain Taylor polynomals to them when doing linearization with them). Thanks for all your input. – Eleven-Eleven May 06 '21 at 16:27
  • @Eleven-Eleven: thanks for your feedback. I am not in academic profession and my only interaction with students is through this website. My only gripe with the tools like L'Hospital's is that most calculus students think of it as panacea to all limit problems and often don't really know how the rule works. In fact if one understands the Taylor series or L'Hospital's Rule, I am sure they would not ask this kind of limit question here as it would already be a cakewalk for them. – Paramanand Singh May 06 '21 at 16:29
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The expression seems designed for sheer cussedness if one attempts many of the common methods, since the indices of the roots differ so the "conjugate-factor" method is not much of an option. (And one should avoid applying LHR to expressions with sums or differences of radicals unless one enjoys suffering...)

An approximation of some sort is more helpful. Ross Millikan's approach makes use of the "binomial approximation". Another way that is accessible, since roots of suitable polynomials are being taken, is to "complete" the cubes and squares:

$$ \lim_{n \ \rightarrow \ +\infty} \ \sqrt[3]{8n^3+4n^2+n+11} \ - \ {\sqrt{4n^2+n+9}} $$ $$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{\left( n^3 \ + \ \frac12 \ n^2 \ + \ \frac18 \ n \ + \ \frac{11}{8} \right)} \ - \ 2 \ {\sqrt{\left(n^2 \ + \ \frac14 \ n \ + \ \frac94 \right)}} $$ $$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{ \ \left[ n^3 \ + \ 3·\frac16 \ n^2 \ + \ 3·\frac{1}{36} \ n \ + \ \frac{1}{216} \right]+ \ \left(\frac18 - \frac{1}{12} \right) n \ + \ \left(\frac{11}{8} - \frac{1}{216} \right)} \ - \ 2 \ \sqrt{\left[n^2 \ + \ 2·\frac18 \ n \ + \ \frac{1}{64} \right] \ + \ \left( \frac94 - \frac{1}{64} \right)} $$ $$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{ \ \left( n \ + \ \frac16 \right)^3 \ + \ \left[ \ \text{small terms} \ \right]} \ \ - \ \ 2 \ \sqrt{ \ \left(n \ + \ \frac18 \right)^2 \ + \ \left[ \ \text{small terms} \ \right]} $$

$$ \approx \ \lim_{n \ \rightarrow \ +\infty} \ \ 2· \left( n \ + \ \frac16 \right) \ \ - \ \ 2 · \left| \ n \ + \ \frac18 \ \right| \ \ , $$

which agrees with the "binomial approximation" result. (One has to magnify a graph a fair bit to see that the limit is not zero.)

Incidentally, this also shows that the horizontal asymptote for this function is "one-sided". We observe that the asymptote corresponding to the limit "at negative infinity" for this expression is a line of slope $ \ 4 \ \ , $ giving us $$ \lim_{n \ \rightarrow \ -\infty} \ \sqrt[3]{8n^3+4n^2+n+11} \ - \ {\sqrt{4n^2+n+9}} \ \ = \ \ -\infty \ \ . $$ (We can consider this situation since $ \ 4x^2+x+9 > 0 \ $ for all real numbers.)

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Set $\dfrac1n=h,$

$$\lim_{n\to\infty}\sqrt[p]{2^pn^p+a_1n^{p-1}+\cdots+a_{p-1}n+a_p}-2n$$

$$=2\lim_{h\to0}\dfrac{\left(1+\dfrac{a_1}{2^p}h+\cdots+\dfrac{a_{p-1}}{2^p}h^{p-1}+\dfrac{a_p}{2^p}h^p\right)^{1/p}-1}h$$

$$=2\lim_{h\to0}\dfrac{1+\dfrac{a_1}{2^p}h+\cdots+\dfrac{a_{p-1}}{2^p}h^{p-1}+\dfrac{a_p}{2^p}h^p-1}h\cdot\dfrac1{\sum_{r=0}^{p-1}\left(1+\dfrac{a_1}{2^p}h+\cdots+\dfrac{a_{p-1}}{2^p}h^{p-1}+\dfrac{a_p}{2^p}h^p\right)^r}$$

$$=\dfrac{2a_1}{p2^p}$$

So, our required limit will be $$\dfrac{2\cdot4}{3\cdot2^3}-\dfrac{2\cdot1}{2\cdot2^2}=\dfrac13-\dfrac14$$

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$$ \begin{align*} &\lim_{n\rightarrow \infty} \left( \sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9} \right) \\ &=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right) \\ &=\lim_{n\rightarrow \infty} 2n\left( \sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-1 \right) +\lim_{n\rightarrow \infty} 2n\left( 1-\sqrt{1+\frac{1}{4n}+\frac{9}{4n^2}} \right) \\ &=\lim_{n\rightarrow \infty} 2n\cdot \frac{1}{3}\left( \frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3} \right) -\lim_{n\rightarrow \infty} 2n\cdot \frac{1}{2}\left( \frac{1}{4n}+\frac{9}{4n^2} \right) \\ &=\frac{2}{3}\lim_{n\rightarrow \infty} \left( \frac{1}{2}+\frac{1}{8n}+\frac{11}{8n^2} \right) -\lim_{n\rightarrow \infty} \left( \frac{1}{4}+\frac{9}{4n} \right) \\ &=\frac{1}{3}-\frac{1}{4} \\ &=\frac{1}{12} \end{align*} $$

xuke0721
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