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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.2011-02-20
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    Hi ? Please ? Homework ? Any thoughts ?2011-02-20
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    @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.2011-02-20
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    lol..ok i am sorry if i am rude..2011-02-20
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    @white: We, would like to know what you have tried that all.2011-02-20
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    i have tried by multiplying {(〖1+x^2+x^3)〗^(1/3) to get rid the 1/3..but i seems to get bac the answer of 1..as if i substitute infinity on the earlier equation..the answer is the same is equal to 1..so am i doin correctly or....wrong???2011-02-20
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    yup..it is the correct interpretation..but if the answer is 1..i juz substitute the infinity to x..than i get the answer??2011-02-20
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    um, I think the answer should be 1/3, not 1.2011-02-20
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    izit??can u please teach me hw??@Willie Wong♦2011-02-20
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    The answer is $\frac{1}{3}$. Unless you use either Taylor series or L'Hospital's rule, it will be a bit messy. (But is doable with the difference of cubes identity and such)2011-02-20
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    In general, $\lim_{x\rightarrow\infty} \sqrt[n]{x^n+ax^{n-1}+O(x^{n-2})}-x=\frac{a}{n}$.2011-02-20

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