Calculating limit of a square root function, in terms of N and epsilon

Question

I've been trying to solve this question:

Prove directly from the definition of a limit, in terms of $N$ and $\varepsilon$:

$$\lim_{n\to\infty} \sqrt{\frac{4n+1}{n}} =2$$

I'm not supposed to use any known properties and use the definition directly.

So, in order to prove this statement, we need to prove that for every $\varepsilon > 0$, there is a natural $N$, which for every $n>N$, the equation $\left|\sqrt{\frac{4n+1}{n}} - 2\right| < \varepsilon$

And since $\sqrt {\frac{4n+1}{n}} = \sqrt {4+\frac{1}{n}} > 2$ for every $n$, we can claim:

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

$$=\sqrt{4+\frac{1}{n}} - \sqrt 4 < \varepsilon$$

$$= \frac{\left(\sqrt{4+\frac{1}{n}} - \sqrt4\right)\left(\sqrt{4+\frac{1}{n}} + \sqrt4\right)}{\left(\sqrt{4+\frac{1}{n}} + \sqrt4\right)}< \varepsilon$$

$$=\frac{4+\frac{1}{n}-4}{\sqrt{4+\frac{1}{n}}+\sqrt4} < \varepsilon$$

$$= \frac{1}{n} \cdot \frac{1}{\sqrt{4+\frac{1}{n}}+\sqrt4}$$

I'm not sure where to go on from here in order to find an $N$. I'd appreciate some help, thank you!

Answer 1

Note that$$0<\sqrt{\frac{4n+1}n}-2=\frac{\frac1n}{\sqrt{\frac{4n+1}n}+2}<\frac1{2n}.$$So, given $\varepsilon>0$. Take $N\in\Bbb N$ such that $N>\frac2\varepsilon$, and then$$n\geqslant N\implies\left|\sqrt{\frac{4n+1}n}-2\right|<\frac1{2n}\leqslant\frac1{2N}<\varepsilon.$$

Answer 2

You have $b_n=\sqrt{4+\frac{1}{n}}$. Let us define sequences $a_n=\sqrt{4}$, $c_n=\sqrt{\left(2+\frac{1}{n}\right)^2}=\sqrt{4+\frac{4}{n}+\frac{1}{n^2}}$. Evidently $a_n<b_n<c_n~n\in\mathbb{N}$, $a_n \to 2$ and also $c_n \to 2$. Consequently $b_n \to 2$.

Answer 3

Alternative approach

$\sqrt{4 + 1/n} < 2 + \epsilon = \sqrt{4 + 4\epsilon + \epsilon^2}.$

Therefore, you have to choose $n$ such that $(4 + 1/n) < (4 + 4\epsilon + \epsilon^2)$.

This means that $(1/n) < 4\epsilon + \epsilon^2.$

This means that $n > \frac{1}{\epsilon} \times \frac{1}{4 + \epsilon}.$

Since $\frac{1}{4 + \epsilon} < 1$, you can simply specify:
$N > \frac{1}{\epsilon} > \frac{1}{\epsilon} \times \frac{1}{4 + \epsilon}.$

Then, any positive integer $n > N$ will also be satisfactory.