- #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)

##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.