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Calculate: $\lim _{n\rightarrow \infty }{n}^{3/2}[\sqrt {{n}^{3}+3}-\sqrt {{n}^{3} -3}]$

What do the brackets mean? I know sometimes they are used to denote a function that returns only the integer part of a number, like $f(x) = [x]$ has values of $0$ on $(0,1)$ and then jumps to $1$ on [1,2) and then $2$ on $[2,3)$ and so on... Is this what is meant here?

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If the square brackets meant integer part, as you speculate, the answer would be fairly easily $0$. I am reasonably confident that you are expected to think of them as a variant of ordinary parentheses. If you want, you can cover all possible bases by adding a remark about what happens if one interprets the square brackets as meaning integer part. – André Nicolas Feb 2 '12 at 20:05

3 Answers 3

up vote 2 down vote accepted

Here the brackets are equivalent to $($ $)$. I am saying so because we don't usually use the integer part function in a calculus or analysis context. If you are in a number theory context, then those might mean the integer part function, but then I don't see why you would be computing this limit.

Hope that helps,

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We don't use integer part in calculus or analysis context? Really? – Aryabhata Feb 2 '12 at 20:02
Limits are sometimes important in number theory, too. – Jonas Meyer Feb 2 '12 at 20:16
I looked at the question as an exercise in a course. Don't you agree now? – Patrick Da Silva Feb 2 '12 at 20:21
thanks, I was getting no where with integer-part. To all you professors and teachers out there, remember that a small typo or brackets instead of regular parentheses can mean hours of confusion for your students. (I took this from a previous test so maybe the professor was there to explain that these are just ordinary parenthesis and were miss-typed for some reason) – nofe Feb 2 '12 at 20:24
Patrick: I do agree that it is equivalent to $()$ in this context. I don't agree that we don't use integer part in calculus or analysis context. I have personally assigned limit problems that involve the floor function, and they show up on this site sometimes too (many of them likely exercises in courses). @nofe: Thank you for the advice. I prefer $\lfloor x\rfloor$ to $[x]$ anyway for the floor function. – Jonas Meyer Feb 2 '12 at 22:32

It is unlikely to be integer part, as that would make the $n^{3/2}$ term irrelevant.

I am quite sure that it is the normal brackets: ().

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Or, using the binomial theorem, ${n}^{3/2}(\sqrt {{n}^{3}+3}-\sqrt {{n}^{3}-3}) = n^3(\sqrt {1+3/n^3}-\sqrt {1-3/n^3}) $ $ = n^3((1 + 3/(2n^3) + O(1/n^6)) - (1 - 3/(2n^3) + O(1/n^6)) $ $ n^3(3/n^3 + O(1/n^6) = 3 + O(1/n^3) $.

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