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Find $$ \lim_{x \to \infty} \sqrt[3]{1 + x^{2} + x^{3}} -x$$

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@white: I have edited the question, please if this is what you wanted to post or not. – anonymous Feb 20 '11 at 13:39
Hi ? Please ? Homework ? Any thoughts ? – Sam Feb 20 '11 at 13:39
@white: As Arturo, always keeps mentioning, please pose the question in a more polite form. This is like asking a homework question. We would like to know what you have tried and where you are finding difficulty. – anonymous Feb 20 '11 at 13:40
lol..ok i am sorry if i am rude.. – user7284 Feb 20 '11 at 13:43
In general, $\lim_{x\rightarrow\infty} \sqrt[n]{x^n+ax^{n-1}+O(x^{n-2})}-x=\frac{a}{n}$. – Eric Naslund Feb 20 '11 at 23:59

HINT $\ $ It's simply a first derivative: $\: $ changing variables $\rm\ x\to 1/x\ $ transforms it to

$$\rm\displaystyle\ \lim_{x\to\ 0^{+}}\ \frac{f(x)-f(0)}x\ =\ f\:'(0) \ \ \ for\ \ \ f(x) = (1+x+x^3)^{1/3}$$

Now it is easy to calculate $\rm\ f\:'(0)\ =\ 1/3\ $ by direct evaluation (it's not indeterminate). Namely

$$\rm f\:'(x)\ =\ \frac{d}{dx}\ (1+x+x^3)^{1/3}\ =\ \frac{1+3\ x^2}{3\ (1+x+x^3)^{2/3}}\ \ \Rightarrow\ \ f\:'(0)\ =\ \frac{1}3$$

Note that this method employs only knowledge of the definition of the derivative and basic rules for calculating derivatives of polynomial and powers. It does not employ more advanced techniques such as L'Hospital's rule, or (binomial) power series expansions, etc.

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is this lopital?? – user7284 Feb 20 '11 at 19:29
@white: The above easy calculation of $\rm\ f\:'(0)\ $ doesn't use L'Hopital's rule (it's not indeterminate). While there is indeed a close relationship between L'Hospital's rule, derivatives, Taylor's formula, the mean-value-theorem, etc., that doesn't imply that use of the latter implies use of L'Hospital's rule. The above answer uses only the definition of the derivative and the basic rules for calculating derivatives. – Bill Dubuque Feb 20 '11 at 19:53
I like this answer, +1. To explain something white, this actually is L'hopitals rule, but not quite. I say not quite because Bill's argument actually shows that L'Hopitals Rule follows immediately from the definition of the derivative when the denominator is $x$. – Eric Naslund Feb 20 '11 at 22:41

Here is a more elementary way using the difference of cubes identity. (It is not as elegant as the Taylor series presented by Willie Wong, but requires less background)

Since $x=\sqrt[3]{x^{3}}$, we are looking at $\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}$.Recall that the cubic identity $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$, which tells us that

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

and hence


Divide the top and bottom of the left hand side by $x^{2}$ to find $$\frac{1+\frac{1}{x^{2}}}{\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$

Since $$\lim_{x\rightarrow\infty}1+\frac{1}{x^{2}}=1$$ and $$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1=3$$ we see by the quotient rule for limits $$\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}=\frac{1}{3}.$$

Hope that helps,

Edit: This faq question: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ was made shortly after this post to help answer it more generally.

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hey..thx for the info..but unfortunately i'm not so understand about this method..abit confusing as from the 3rd part of ur solution,the x has been forfited n i duno where to cancel it off..sry bro i not so good in maths..XD – user7284 Feb 20 '11 at 17:19
Which part exactly? I might be able to explain better. – Eric Naslund Feb 20 '11 at 22:43

hint: you can rewrite the terms inside the limit as

$$ x \cdot \left( \sqrt[3]{\frac{1}{x^3} + \frac{1}{x} + 1} - 1\right) $$

for the term underneath the cube-root, use the Taylor expansion of the cube-root function near the value 1:

$$ \sqrt[3]{1 + y} = 1 + \frac{1}{3}y - \frac{1}{9} y^2 + \ldots $$

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bionomial???ok..i try it nw..thx for ur help.. – user7284 Feb 20 '11 at 14:32
ermm,@Willie Wong♦ ,i am not so sure about the taylor expansion as u mention..can u explain more on hw to apply it on the cube roots?? – user7284 Feb 20 '11 at 14:41
The Taylor expansion of (1+x)^k starts with 1+kx, even if k is not an integer. For this problem you only need the $y/3$ term. – Ross Millikan Feb 20 '11 at 16:39
You may want to take a look at to refresh your memory on the Taylor/binomial expansion for (1+x)^k. – Willie Wong Feb 20 '11 at 17:45

A very similar question was asked recently here...

The limit follows immediately upon showing, using the mean value theorem, that $$ \sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3 }} - \sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3 - \bigg(\frac{x}{3} - \frac{{26}}{{27}}\bigg)}} \to 0 $$ as $x \to \infty$.

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I hope you understand where the term $(x + \frac{1}{3})^3 - (\frac{x}{3} - \frac{{26}}{{27}})$ comes from. – Shai Covo Feb 20 '11 at 19:02
nope..i am noob – user7284 Feb 20 '11 at 19:28
@white: It is equal to $x^3 + x^2 +1$. – Shai Covo Feb 20 '11 at 19:37

For such problems, quite often it pays off to use the identity

$a^b = \exp[ b * \log(a)]$

because then one can apply all his/her knowledge about exp and log.

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Can you explain? I don't see how you use this for this example unless you expand $e^{b\log a}=1+b\log a+\frac{1}{2}(b\log a)^2+\cdots$ and then expand the $\log$'s into their power series as well. Am I missing something? – Eric Naslund Feb 20 '11 at 16:38
The comment was for an earlier version of the question which looked quite a bit different [$\lim_{x\to \infty} (1 + x^2 + x^3)^{1/3 -x}$]. – Fabian Feb 20 '11 at 20:27

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