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Need to solve this very simple limit $$ \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$

I know how to solve these limits: by using $a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - very tedious, boring and tiring. I hope there is some artful and elegant solution. Thank you!

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  • $\begingroup$ What about the binomial expansion? $\endgroup$
    – Kartik
    Sep 13 '15 at 7:51
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    $\begingroup$ I would say the standard way (not by using L'Hospital's rule) to solve this limit is not very tedious, boring and tiring. In the expression for 'a-b' divide the numerator and denominator of 'x' and the result is obvious. $\endgroup$
    – georg
    Sep 13 '15 at 8:05
  • $\begingroup$ @georg but not so elegant as below :)) $\endgroup$ Sep 13 '15 at 9:40
  • $\begingroup$ Apparently it will ;-) $\endgroup$
    – georg
    Sep 13 '15 at 13:51
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$$\lim _{ x\to \infty \: } \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) =$$ $\lim _{ x\to \infty \: } \frac { \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } =$$$\ =\lim _{ x\to \infty : } \frac { 3x^{ 2 }+4x+1-3x^{ 2 }-9x-2 }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } =\lim _{ x\rightarrow \infty }{ \frac { -5x-1 }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } } $$ now ,the power of denominator of polynomial is higher than the numerator so the limit is equal to $0$

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    $\begingroup$ This is the best way to calculate the limit. $\endgroup$
    – A.Γ.
    Sep 13 '15 at 9:13
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You have $$f(x)=\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}=\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}+\frac{1}{3x^2}}-\sqrt[3]{1+\frac{3}{x}+\frac{2}{3x^2}}\right)$$ Using Taylor expansion at order one of the cubic roots $\sqrt[3]{1+y}=1+\frac{y}{3}+o(y)$ at the neighborhood of $0$, you get: $$f(x)=\sqrt[3]{3x^2}\left(\frac{4}{9x}-\frac{1}{x}+o\left(\frac{1}{x}\right)\right)$$ hence $$\lim\limits_{x \to \infty} f(x)=0$$

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  • $\begingroup$ Thanks! Can you explain me plese your second expression, where you using Taylor expansion? $\endgroup$ Sep 13 '15 at 9:01
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    $\begingroup$ Yes I was using Taylor expansion and I edited the post to provide more details. $\endgroup$ Sep 13 '15 at 9:36
  • $\begingroup$ Oh, now I am in clear! Thanks!! $\endgroup$ Sep 13 '15 at 9:42
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You can use the binomial theorem to expand this.

$$\lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right)$$

Here 1 and 2 are the smallest terms and they can be ignored.

\begin{align} &\lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) \\=& \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x}-\sqrt[3]{3x^2+9x}\right) \\=& \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2\left(1+\frac{4}{3x}\right)}-\sqrt[3]{3x^2\left(1+\frac3x\right)}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left(\sqrt[3]{1+\frac{4}{3x}}-\sqrt[3]{1+\frac3x}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left({1+\frac{4}{9x}}-{1-\frac1x}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3x^2}\left({\frac{4-9}{9x}}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3}x^{2/3}\left({\frac{-5}{9x}}\right) \\=& \lim _{x\to \infty \:}\sqrt[3]{3}x^{-1/3}\left({\frac{-5}{9}}\right) \\=& \,0 \end{align}

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    $\begingroup$ Not clear why you can ignore constants and cannot ignore linear terms which are also of smaller order. $\endgroup$
    – A.Γ.
    Sep 13 '15 at 9:05
  • $\begingroup$ @A.G. Hm. Can we ignore constants? $\endgroup$ Sep 13 '15 at 9:39
  • $\begingroup$ @PersonaNonGrata Well, in this particular case yes, but the explanation why it is possible gonna cost you as much efforts as to solve the problem without ignoring anything. $\endgroup$
    – A.Γ.
    Sep 13 '15 at 9:44
  • $\begingroup$ @A.G. Oh. So better don't ignore constants? :)) $\endgroup$ Sep 13 '15 at 9:46
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    $\begingroup$ @PersonaNonGrata Better is to do first the Taylor expansion of the original function and then ignore properly lower order terms. $\endgroup$
    – A.Γ.
    Sep 13 '15 at 9:52
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Actually, this question can be solved without calculation if you're familiar with power.

If the first term(in the square root), the biggest term is $3x^2$ and so does the second term. And any term has power below 2 are negligible. So you can just forget them and get your answers, which is $0$.

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    $\begingroup$ "Any term of lower order is negligible" is not correct. Example: $\sqrt{x^2+x+1}-\sqrt{x^2-x+1}$. Neglecting lower order terms gives the wrong limit $0$ at infinity. (Correct limit is $1$ in my example). $\endgroup$
    – A.Γ.
    Sep 13 '15 at 9:08
  • $\begingroup$ @A.G. Yes. I am wrong. Thanks for pointing it out. To serve as a counter-example, I'd better remain it here instead edit it or delete it. $\endgroup$
    – Rowan
    Sep 13 '15 at 10:41
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Can you help with O-symbols? It's all right here? $$f(x) = \sqrt[3]{3x^2}\left(1 + \frac{4}{9x} + O\left(\frac{1}{x^2}\right) - 1 - \frac{1}{x} -O \left(\frac{1}{x^2}\right)\right)= \sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right). $$

Hence $$\lim _{x\to \infty }\sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right)= \lim_{x \to \infty} \sqrt[3]{3}{x^{-1/3}}^{\to 0} \left( - \frac{5}{9}+\frac{1}{18x}^{\to 0} \right) = 0.$$

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