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Evaluate the limit

$\lim_{x\rightarrow \infty}(\sqrt[3]{x^3+x^2}-x)$

I know that the limit is $1/3$ by looking at the graph of this function, but I struggle to show it algebraically.

Is there anyone who can help me out and maybe even provide a solution to this problem?

David C. Ullrich
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fredericoamigo
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3 Answers3

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You have $$\sqrt[3]{x^3+x^2} - x = x \left(\sqrt[3]{1 + \frac{1}{x}} - 1 \right) = x \left( 1 + \frac{1}{3x} + o\left(\frac{1}{x} \right) - 1\right) = \frac{1}{3} + o(1)$$

and you are done.

TheSilverDoe
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  • Thanks! Clean solution! I haven't seen that before. Is it the binominal formula? If so om not sure what the $o$ stands for. Also. How do you fit this into the limit notation ($\lim_{x\rightarrow \infty$}. – fredericoamigo Oct 20 '21 at 10:39
  • @fredericoamigo See https://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation – Gary Oct 20 '21 at 10:41
  • @fredericoamigo If you have not learned Taylor expansions yet, then maybe my answer is a bit difficult to understand. The idea is to make polynomial approximation of functions to determine the limits. The approximation is contained in the "little $o$ notation", as you can see in Gary's link. You can see David C. Ullrich's answer to get a more elementary answer. – TheSilverDoe Oct 20 '21 at 10:43
  • @TheSilverDoe: I know some basic Taylor series expressions, but I first encountered them a week ago, so I am still getting comfortable with them. Is the $o$ simply $R_nf(x)$? – fredericoamigo Oct 20 '21 at 10:54
  • @TheSilverDoe: Also: how did you factor out x in $x(\sqrt[3]{1+\frac{1}{x}}-1)$? – fredericoamigo Oct 20 '21 at 11:00
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    @fredericoamigo The factorization is the "easy" part : just write $$\sqrt[3]{x^3+x^2} =\sqrt[3]{x^3 \left( 1 + \frac{1}{x}\right)} = \sqrt[3]{x^3} \sqrt[3]{\left( 1 + \frac{1}{x}\right)} = x \sqrt[3]{ \left( 1 + \frac{1}{x}\right)} $$ – TheSilverDoe Oct 20 '21 at 15:17
  • @TheSilverDoe: Thanks! I have actually solved it using another method (which I'm about to post in the near future). That factorization was actually the critical part of my solution:-) In other words, you have been very helpful! – fredericoamigo Oct 20 '21 at 22:40
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Hint: If $t=1/x$ then $$\sqrt[3]{x^3+x^2}-x=\frac{\sqrt[3]{1+t}-1}{t}$$

David C. Ullrich
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  • Thank you help. Unfortunately, I'm not familiar with the $t=1/x$ trick you are referring to. Can you elaborate? – fredericoamigo Oct 20 '21 at 10:42
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    @fredericoamigo You go from $x\to +\infty$ to $t \to 0$. – Gary Oct 20 '21 at 10:46
  • @Gary: Ok. This might be a stupid question, but where did you get the $t$ from? (This is not a method I have encountered before in the books I've used so far) - so I would love to learn it and understand it. – fredericoamigo Oct 20 '21 at 10:52
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    @fredericoamigo See the answers here: https://math.stackexchange.com/questions/1069642/finding-a-limit-using-change-of-variable-how-come-it-works – Gary Oct 20 '21 at 10:53
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Use the equality $a^3 - b^3 = (a - b)(a^2 + a b + b^2)$ for $a = \sqrt[3]{x^3 + x^2}$ and $b = x$. Convert your expression to $$\frac{(a - b)(a^2 + a b + b^2)}{a^2 + a b + b^2}$$ and simplify further.

Antoine
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  • Thanks! I have attempted that for several hours now, and for some reason, I get completely stuck. I would love to be able to learn this solution. Do you think you can show me how this is done step by step? If so I would be extremely grateful. – fredericoamigo Oct 20 '21 at 10:48