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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!

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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
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5 Answers 5

up vote -1 down vote accepted

Proof by definition:

Let $\epsilon>0$ given.

Looking on:

$$|\sqrt{4n^2+n}-2n-\frac{1}{4}|<\epsilon$$

We can see that $\sqrt{4n^2+n}-2n-\frac{1}{4}>0$ for some $n>N$.

Therefore we left with

$$\sqrt{4n^2+n}-2n-\frac{1}{4}<\epsilon$$

or

$$\sqrt{4n^2+n}<2n+\frac{1}{4}+\epsilon$$

Taking square from both sides

$${4n^2+n}<4n^2+\frac{1}{16}+{\epsilon}^2+n+4n\epsilon+\frac{\epsilon}{2}$$

which is

$$\frac{1}{16}+{\epsilon}^2+4n\epsilon+\frac{\epsilon}{2}>0$$

or,

$$4n\epsilon>\frac{1}{16}-{\epsilon}^2-\frac{\epsilon}{2}$$

dividing by $4\epsilon$,

$$n>\frac{1}{16\epsilon}-{\epsilon}-\frac{1}{2}$$

Now, by choosing:

$$N=[\frac{1}{16\epsilon}-{\epsilon}-\frac{1}{2}]$$

we are done.

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OMG this is so clear...thanks a lot!! One question I want to ask is that we can get rid of the absolute value because $\sqrt{4n^2+n} > 2n-\frac{1}{4}$?? –  eChung00 Sep 4 '13 at 2:11
    
Yes, its true for any $n\geq 0 $ (If I not mistaken). –  Salech Alhasov Sep 4 '13 at 2:16
    
Thanks a lot!! It was great help!! –  eChung00 Sep 4 '13 at 2:18
    
You are welcome! –  Salech Alhasov Sep 4 '13 at 2:19
    
I can't follow this at all! Your sentence "We can see that..." seems to be false for all $n$, because $\sqrt{4n^2+n} < \sqrt{4n^2+n+\frac1{16}} = 2n+\frac14$. –  TonyK Nov 24 '13 at 13:33
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$$\frac{\sqrt{4n^2+n}-2n}{1}=\frac{\left(\sqrt{4n^2+n}-2n\right)\left(\sqrt{4n^2+n}+2n\right)}{\sqrt{4n^2+n}+2n}=\frac{4n^2+n-4n^2}{\sqrt{4n^2+n}+2n}$$

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
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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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