Find $$ \lim_{x \to \infty} \sqrt[3]{1 + x^{2} + x^{3}} -x$$
Limit of algebraic function avoiding l'Hôpital's rule
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analysis
limits
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1@white: I have edited the question, please if this is what you wanted to post or not. – 2011-02-20
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0Hi ? Please ? Homework ? Any thoughts ? – 2011-02-20
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1@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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0lol..ok i am sorry if i am rude.. – 2011-02-20
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0@white: We, would like to know what you have tried that all. – 2011-02-20
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0i 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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0yup..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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0um, I think the answer should be 1/3, not 1. – 2011-02-20
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0izit??can u please teach me hw??@Willie Wong♦ – 2011-02-20
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0The 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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3In general, $\lim_{x\rightarrow\infty} \sqrt[n]{x^n+ax^{n-1}+O(x^{n-2})}-x=\frac{a}{n}$. – 2011-02-20