# Simple limit of a sequence

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!

• What about the binomial expansion? Sep 13 '15 at 7:51
• 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. Sep 13 '15 at 8:05
• @georg but not so elegant as below :)) Sep 13 '15 at 9:40
• Apparently it will ;-) Sep 13 '15 at 13:51

$$\lim _{ x\to \infty \: } \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) =$$ $\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$0$• This is the best way to calculate the limit. – A.Γ. Sep 13 '15 at 9:13 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 $$\lim\limits_{x \to \infty} f(x)=0$$ • Thanks! Can you explain me plese your second expression, where you using Taylor expansion? Sep 13 '15 at 9:01 • Yes I was using Taylor expansion and I edited the post to provide more details. Sep 13 '15 at 9:36 • Oh, now I am in clear! Thanks!! Sep 13 '15 at 9:42 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} • Not clear why you can ignore constants and cannot ignore linear terms which are also of smaller order. – A.Γ. Sep 13 '15 at 9:05 • @A.G. Hm. Can we ignore constants? Sep 13 '15 at 9:39 • @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. – A.Γ. Sep 13 '15 at 9:44 • @A.G. Oh. So better don't ignore constants? :)) Sep 13 '15 at 9:46 • @PersonaNonGrata Better is to do first the Taylor expansion of the original function and then ignore properly lower order terms. – A.Γ. Sep 13 '15 at 9:52 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 is3x^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$0$. • "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).
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 $$\lim _{x\to \infty }\sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right)= \lim_{x \to \infty} \sqrt[3]{3}{x^{-1/3}}^{\to 0} \left( - \frac{5}{9}+\frac{1}{18x}^{\to 0} \right) = 0.$$