Prove $\,\lim\limits_{n\to\infty} \sqrt{\frac{4n+1}{n}}=2$

Question

As shown in the title, I'm asked to prove $\lim\limits_{n\to\infty} \sqrt{\dfrac{4n+1}{n}}=2$ using the $\epsilon,N$ definition only.
My solution was completely different from the solution they posted so I want to know if I did something wrong.
Let $\epsilon>0$, we have to find $N$ s.t $\forall n>N$. $\left|\sqrt{\dfrac{4n+1}{n}}-2\right|<\epsilon$. Since $n$ is iterating over naturals, we can say that $\sqrt{\dfrac{4n+1}{n}}-2$ is always positive by plugging in the minimum of $\mathbb{N}$.

I continued by multiplying by $\left(\sqrt{\dfrac{4n+1}{n}}+2\right)$.\ $\dfrac{\left(\sqrt{\dfrac{4n+1}{n}}-2\right)\left(\sqrt{\dfrac{4n+1}{n}}+2\right)}{\left(\sqrt{\dfrac{4n+1}{n}}+2\right)}= \dfrac{\dfrac{4n+1}{n}-4}{\left(\sqrt{\dfrac{4n+1}{n}}+2\right)}\stackrel{(*)}{<}\dfrac{1}{n}$ while (*) was using the previous conclusion that the denominator is always positive and greater than $1$.

Now I can say that we can take $N=\dfrac{1}{\epsilon}+10$ and this satisfies the desired.

Answer 1

An idea:

$$\frac{4n+1}n-4=\frac1n\implies\left|\frac{4n+1}n-4\right|=\left|\frac1n\right|=\frac1n$$

You can already from here deduce $\;\frac{4n+1}n\xrightarrow[n\to\infty]{}4\;$ by means of a $\;\delta,\epsilon\;$ argument or however, and

since the sequence is clearly positive all the time, we get

$$\sqrt{\frac{4n+1}n}\xrightarrow[n\to\infty]{}\sqrt4=2$$

Or else:

$$\left|\sqrt{\frac{4n+1}n}-2\right|=\left|{\frac{\sqrt{4n+1}}{\sqrt n}}-\sqrt 4\right|=\left|\frac{\sqrt{4n+1}-\sqrt{4n}}n\right|=\left|\frac1{n\left(\sqrt{4n+1}+\sqrt{4n}\right)}\right|\le$$

$$\le\frac1{2n\sqrt{4n}}=\frac1{4n^{3/2}}$$

and again: a $\;\delta,\epsilon\;$ argument or whatever.

Answer 2

Your proof looks fine. Only problem is your reasoning why we can leave out absolute value. $\sqrt{\frac{4n+1}{n}}-2$ has no minimum value. It's easier to see in form $\sqrt{4+\frac{1}{n}}-2$ where $\sqrt{4+\frac{1}{n+1}}-2$ is a smaller number. My approach to show that the sequence is positive for all n would be: $2=\sqrt{4}<\sqrt{4+\frac{1}{n}}$ (because $\frac{1}{n}>0$)$\iff0<\sqrt{4+\frac{1}{n}}-2$.

Answer 3

My approach.

For any $\,\varepsilon>0\,$ there exists $\,N_{\varepsilon}=\left\lfloor\dfrac1{\varepsilon(\varepsilon+4)}\right\rfloor+1\in\Bbb N\,$ such that for any $\,n>N_{\varepsilon}\,$ it results that

$\left|\sqrt{\dfrac{4n+1}n}-2\right|=\left|\sqrt{4+\dfrac1n}-2\right|=\sqrt{4+\dfrac1n}-2<$

$<\sqrt{4\!+\!\dfrac1{N_{\varepsilon}}}-2=\!\sqrt{4\!+\!\dfrac1{\left\lfloor\!\frac1{\varepsilon(\varepsilon+4)}\!\right\rfloor\!+\!1}}-2\!\!\!\underset{\overbrace{\text{ because }\left\lfloor\!\frac1{\varepsilon(\varepsilon+4)}\!\right\rfloor+1>\frac1{\varepsilon(\varepsilon+4)}}}{<}$

$<\sqrt{4+\dfrac1{\frac1{\varepsilon(\varepsilon+4)}}}-2=\sqrt{4+\varepsilon\big(\varepsilon\!+\!4\big)}-2=$

$=\sqrt{\varepsilon^2+4\epsilon+4}-2=\sqrt{\big(\varepsilon+2\big)^{\!2}}-2=\varepsilon\,.$

Answer 4

I thought that it might be instructive to present a more streamlined approach. To that end we proceed.

First, we can write the sequence of interest as

$$\sqrt{\frac{4n+1}{n}}=2\sqrt{1+\frac1{4n}}\tag1$$

Second, we observe that

$$1<\sqrt{1+\frac1{4n}}\le 1+\frac1{8n}\tag2$$

Using $(1)$ and $(2)$ together reveals the estimates

$$2<\sqrt{\frac{4n+1}{n}}<2+\frac1{4n}$$

which imply that

$$0<\sqrt{\frac{4n+1}{n}}-2<\frac1{4n}\tag 3$$

And now, from $(3)$, for any $\varepsilon>0$, $\left|\sqrt{\frac{4n+1}{n}}-2\right|<\varepsilon$ whenever $n>\frac1{4\varepsilon}$. And we are done!