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

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In summary, the conversation discusses an epsilon-delta exercise where the goal is to show that ##\big| [\sqrt{4n^2 +n} - 2n] - \frac{1}{4}\big|## can be made smaller than any given epsilon. The discussion includes a proof using limit properties and an alternate approach using the basic property of real numbers. The author also mentions the use of the epsilon-delta method in calculus proofs.
  • #1
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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.
 
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  • #2
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.
 
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  • #3
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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  • #4
Delta2 said:
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.
 
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  • #5
Hall said:
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.

Hall said:
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##
 
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  • #6
Mark44 said:
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)##
 
  • #7
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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  • #8
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##.
 
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  • #9
PeroK said:
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?
 
  • #10
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.
 

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

1. What is the definition of a limit?

A limit is the value that a function approaches as the input value approaches a certain value or infinity.

2. How do you prove a limit using the epsilon-delta definition?

To prove a limit using the epsilon-delta definition, you must show that for any epsilon (a small positive number), there exists a delta (a small positive number) such that when the input value is within delta units of the limit, the output value is within epsilon units of the limit.

3. What is the limit of a sequence?

The limit of a sequence is the value that the terms of the sequence approach as the index of the terms approaches infinity.

4. How do you prove a limit of a sequence using the definition?

To prove a limit of a sequence using the definition, you must show that for any epsilon (a small positive number), there exists a natural number N such that when the index of the terms is greater than N, the terms of the sequence are within epsilon units of the limit.

5. How can you use algebraic manipulation to prove a limit?

You can use algebraic manipulation to prove a limit by rearranging the expression and factoring out common terms to simplify the expression and show that it is equivalent to the limit expression.

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