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$$\lim_{n\rightarrow \infty}\left[{\sqrt{4n^2+n}-2n}\right]=\frac{1}{4}$$ I am trying to use the definition of the limit but have no idea how to simplify the expression with radical!


so by the definition, $\forall n > N \rightarrow |a_n - a | < \epsilon, \text{where}\, \ a_n=\sqrt{4n^2+n}-2n\ \, \text{and}\,\ a = \frac{1}{4}$

so after multiply the conjugate and negate $\frac{1}{4}$, I get $\frac{2n-\sqrt{4n^2+n}}{4(\sqrt{4n^2+n}+2n)}$

Since there is still radical in the numerator, I think I have multiply the conjugate again...right?? And then find some formula that is greater!!??? confuse me this real analysis!!

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You forgot to specify $n_0$ in $\lim_{n\rightarrow n_0}$ – par Sep 4 '13 at 0:36
$$\frac{\sqrt{4n^2+n}-2n}{1}=\frac{(\sqrt{4n^2+n}-2n)(\sqrt{4n^2+n}+2n)}{(\sqrt{‌​4n^2+n}+2n)}=\frac{4n^2+n-4n^2}{(\sqrt{4n^2+n}+2n)}$$ Can you continue? – Inquest Sep 4 '13 at 0:39
Multiply by the conjugate over itself, and then divide by n on the top and bottom. – user84413 Sep 4 '13 at 0:39
@Inquest: That was fine as an answer; I was just about to upvote it. – Brian M. Scott Sep 4 '13 at 0:40
@BrianM.Scott, I thought it was more of a comment. I undeleted it. – Inquest Sep 4 '13 at 0:41


Can you continue?

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Hint: $$ \sqrt{4n^2+n}-2n=\frac{n}{\sqrt{4n^2+n}+2n} $$

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Here is a proof I like but is not too easy to generalize:

Let $f(x) = \sqrt{4+x}$. Note that $f(0) = 2$. Now $\lim_{x \rightarrow 0}(\frac{f(x) -f(0)}{x}) = f'(0) = \frac{1}{2\sqrt{4+0}} = \frac{1}{4}$. But also $\lim_{x \rightarrow 0}(\frac{f(x) -f(0)}{x}) = \lim_{x \rightarrow 0}(\frac{\sqrt{4+x}-2}{x})$ by the definition of $f(x)$, and this is also equal to $\lim_{x \rightarrow 0}(\sqrt{\frac{4}{x^2} + \frac{1}{x}} - \frac{2}{x})$. Now when we let $n = \frac{1}{x}$ we recover the original limit $\lim_{n \rightarrow \infty}(\sqrt{4n^2 + n}-2n)$. Hooray!

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The way my professor proved similar question was by using the limit definition of sequence. $\forall n > N \rightarrow |a_n-a| < \epsilon$ and can u explain how did u get the expression $f(x)=\sqrt{4+x}$ – eChung00 Sep 4 '13 at 1:14

I would prove it with Taylor expansion:

$\lim_{n\to\infty} \sqrt{4n^2+n}-2n = \lim_{n\to\infty} 2n(-1+\sqrt{1+\frac{4}{n}})=\lim_{n\to\infty}2n(-1 + 1 +\frac{1}{2}\frac{1}{4n}-\frac{1}{8}(\frac{1}{4n})^2+\dots) = \frac{1}{4}$

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