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

[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)

Spoiler

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.

 

[Mark44]

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, $\lim_{n \to \infty}\sqrt{4n^2 + n} - 2n = \lim_{n \to \infty}\frac 1 {\sqrt{4 + 1/n} + 2} = \frac 1 4$.

 

[Hall]

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.

 

[Mark44]

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 $\epsilon - a < x < \epsilon + a$

 

[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 $(2)\Rightarrow (1)$

 

[PeroK]

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.

 

[PeroK]

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

 

[Hall]

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?

 

[PeroK]

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.