# Proving that ##\lim [\sqrt{4n^2 +n} - 2n] = \frac{1}{4}##

• Hall

#### Hall

Homework Statement
Prove the limit: ##\lim [\sqrt{4n^2 +n} - 2n] = \frac{1}{4}##.
Relevant Equations
I will use epsilon-delta definition.
Discussion: Assume that we can make ##\big| [\sqrt{4n^2 +n} - 2n]- \frac{1}{4}\big| ## to fall down any given number. Given an arbitrarily small ##\varepsilon \gt 0##, we assume
$$\big| [\sqrt{4n^2 +n} - 2n] - \frac{1}{4}\big| \lt \varepsilon$$
$$\big| [\sqrt{4n^2 +n} - 2n]\big| \lt \varepsilon + 1/4$$
Now, we have two problems here, first that we cannot fully isolate ##n## and second ##n## doesn't occur in denominator as in fractional sequences. Without hurting anyone's feelings we would try to solve the second issue first:
## \text{Let's irrationalise the denominator of the expression of the given sequence}##
##\frac{ \left(\sqrt{4n^2 +n} - 2n\right) ~\left( \sqrt{4n^2 +n} + 2n\right)}{ \sqrt{4n^2 +n} +2n}##
##\frac{n }{ \sqrt{4n^2 +n} + 2n} = \frac{1}{ \sqrt{4 +1/n} - 2}##

Let's simply our expression ## \big| [\sqrt{4n^2 +n} - 2n] - \frac{1}{4}\big| = \big| \frac{1}{ \sqrt{4 +1/n} - 2} - 1/4\big| = 1/4 - \frac{1}{ \sqrt{4 +1/n} - 2}~~~\text{ for all} n \in \mathbf{N}##.

Now, it's time for solving the first issue, that is to make ##n## floatable, for that we will estimate our original expression by something bigger than that (this is my official argument, that I claim the following expression to be less that epsilon)
$$1/4 - \frac{1}{ \sqrt{4 +1/n} - 2} \lt 1/4 - \frac{1}{4 +1/n +2} = 1/4 - \frac{1}{6+ 1/n} \lt \varepsilon$$
## 1/4 - \frac{1}{6+ 1/n} \lt \varepsilon ##
##1/4 - \varepsilon \lt \frac{1}{6+1/n} ##
##6/4 - 6 \varepsilon + 1/n ( 1/4 - \varepsilon) \lt 1##
##1/n ( 1/4 - \varepsilon) \lt 6 \varepsilon - 1/2 ##
##n \gt \frac{2 (1/4 - \varepsilon) }{12 \varepsilon -1}##

Formal Proof: (I'm putting it in spoiler so as to help you, otherwise the post will become so lengthy and mobile users will find it hard to scroll such a length)
For any given arbitrarily small ##\varepsilon \gt 0##, take ## N = \frac{2 (1/4 - \varepsilon) }{12 \varepsilon -1}##
##n \gt N \implies n \gt \frac{2 (1/4 - \varepsilon) }{12 \varepsilon -1}##
##1/n \lt \frac{12 \varepsilon -1}{2 (1/4 - \varepsilon)}##
## 1/n ( 1/4 - \varepsilon) \lt 6\varepsilon -1/2##
##6/4 - 6\varepsilon + 1/n(1/4 - \varepsilon) \lt 1 ##
##1/4 - \varepsilon \lt \frac{1}{6 + 1/n}##
##1/4 - \frac{1}{4 + 1/n +2} \lt \varepsilon##
##1/4 - \frac{ 1}{\sqrt{ 4 +1/n} +2 } \lt 1/4 - \frac{1}{4 + 1/n +2} \lt \varepsilon##
##1/4 - \frac{ 1}{\sqrt{ 4 +1/n} +2 } \lt \varepsilon ##
##\big| \frac{ 1}{\sqrt{ 4 +1/n} +2 } - 1/4 \big| \lt \varepsilon##
##\big| [\sqrt{4n^2 +n} - 2n] - 1/4 \big| \lt \varepsilon##

Thus, we can the expression ## \big| [\sqrt{4n^2 +n} - 2n] - 1/4 \big|## to fall below any given number.

So, basically there were three important steps:
1. First get ##n## in denominator.
2. Estimate the original expression by something else, so as to make ##n## to move freely.
3. Judiciously removing the bars of absolute values.

Now, I seek your solemn opinions, gentlemen.

• Delta2

Well that's all very nice but I don't understand your insist on the epsilon-delta proofs. They make the solution of this type of problems. an excercise in solving inequalities which is not exactly the main theme of calculus.

Though ok, I have to say that many proofs of theorems of calculus are done with the epsilon-delta method.

• Hall
Is there a requirement for this problem that the definition of the limit must be used? If not, it's much simpler to use limit properties.
$$\sqrt{4n^2 + n} - 2n = \frac{(\sqrt{4n^2 + n} - 2n)(\sqrt{4n^2 + n} - 2n)}{\sqrt{4n^2 + n} + 2n}$$
$$=\frac n {n(\sqrt{4 + 1/n} + 2} = \frac 1 {\sqrt{4 + 1/n} + 2}$$
This process is called "rationalizing the numerator", not irrationalizing it.
The first and last expressions in the equation above are equal for all n > 0, so they are equal in the limit as n grows arbitrarily large.
Therefore, ##\lim_{n \to \infty}\sqrt{4n^2 + n} - 2n = \lim_{n \to \infty}\frac 1 {\sqrt{4 + 1/n} + 2} = \frac 1 4##.

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• Delta2 and PeroK
Well that's all very nice but I don't understand your insist on the epsilon-delta proofs. They make the solution of this type of problems. an excercise in solving inequalities which is not exactly the main theme of calculus.

Though ok, I have to say that many proofs of theorems of calculus are done with the epsilon-delta method.
Sorry, I didn’t mention that it is an epsilon-delta exercise, occurring in Ross’ Analysis.

• Delta2
Sorry, I didn’t mention that it is an epsilon-delta exercise, occurring in Ross’ Analysis.
I don't have this textbook. Does the fact that the exercise is in Ross mean that it must be done using the limit definition? Note that for this problem ##\delta## is not used.

Given an arbitrarily small
##\varepsilon \gt 0##, we assume
$$\big| [\sqrt{4n^2 +n} - 2n] - \frac{1}{4}\big| \lt \varepsilon$$
$$\big| [\sqrt{4n^2 +n} - 2n]\big| \lt \varepsilon + 1/4$$
This is not quite right. If ##|x - a| < \epsilon##, then ##\epsilon - a < x < \epsilon + a##

• SammyS and Delta2
I don't have this textbook. Does the fact that the exercise is in Ross mean that it must be done using the limit definition? Note that for this problem ##\delta## is not used.

This is not quite right. If ##|x - a| < \epsilon##, then ##\epsilon - a < x < \epsilon + a##
I think for these two inequalities of the OP the second is a consequence of the first but the first is not consequence of the 2nd. In other words ##(1)\Rightarrow(2)## but NOT ##(2)\Rightarrow (1)##

There is an alternative approach. For all ##n## we have ##\sqrt{4n^2+n} < 2n + \frac 1 4##. It is enough to show, therefore, given ##\epsilon## that for large enough ##n## we have ##2n +\frac 1 4 - \epsilon < \sqrt{4n^2 +n}##.

You can work on that.

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• • Hall and Delta2
PS the idea is to use the basic property of real numbers that for ##x,y > 0## we have ##x < y## iff ##x^2 < y^2##.

• Delta2
There is an alternative approach. For all ##n## we have ##\sqrt{4n^2+n} < 2n + \frac 1 4##. It is enough to show, therefore, given ##\epsilon## that for large enough ##n## we have ##2n +\frac 1 4 - \epsilon < \sqrt{4n^2 +n}##.

You can work on that.
The idea is quite interesting, you seem to suggest that at very large ##n## the expression ##\sqrt{4n^2 +n}## is so close to ##2n +1/4## that even a slight deduction in ##2n +1/4## makes it less than ##\sqrt{4n^2 +n}##.

To assure that we can make ##\sqrt{4n^2 +n}## very close to ##2n +1/4##, we will work out for such an ##n##, so, that our claim is true.
$$2n +1/4 -\varepsilon \lt \sqrt{4n^2 +n}$$
##8n +1 - \varepsilon \lt 4 \sqrt{4n^2 +n}##
Squaring both sides and cancelling would yield
##
1+ \varepsilon^2 -3 \varepsilon \lt 16n \varepsilon##
##
\frac{(1+\varepsilon)^2}{16 \varepsilon} \lt n##

So, for ##n## larger than that, the difference between the two expressions become less than ##\varepsilon##.

I got a question for you: how did you come up with such a fine idea that the algebra got reduced to almost nothing?

That's not quite the idea I had in mind:
$$(2n +\frac 1 4 - \epsilon)^2 = 4n^2 + n - 4n\epsilon + (\frac 1 4 - \epsilon)^2$$and then it should be clear.

I don't know where ideas come from. That approach seems to me to capture the essence of the required inequality.