Prove that $\lim_{n\to\infty} \left(\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}\right) =2$ 
Prove that $$\left(\displaystyle\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}\right)_{n \in \mathbb{N}}\to 2$$

My idea was to find some $N$ to do the definition job of convergence.
\begin{align}\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}-2 \right |&=\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})-8n+2}{4n - 1} \right |\\
&=\left |\dfrac{\sqrt{n}+8n-8n+2}{4n - 1} \right |\\
&=\left |\dfrac{\sqrt{n}+2}{4n - 1} \right |\\
&\leq \left |\dfrac{\sqrt{n}+2}{n}\right |\\
&\leq \left |\dfrac{\sqrt{n}}{n}\right |+ \left |\dfrac{2}{n}\right |\\
&\leq \dfrac{1}{\sqrt{n}}+2\\
&\leq 3
\end{align}
so I concluded that the sequence is bounded, so my idea was trying to show that this sequence is monotonous, so I could conclude that it converges. Taking $n = 4k$ I get a subsequence that converges to $2$, so if I knew that the sequence is convergent, the problem would be over.
 A: With what you've done, you don't need to show the sequence is monotonous. To show this converges, we want to say that for all $\varepsilon > 0$ there exists an $N$ such that for all $n>N$
$$\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}-2 \right | < \varepsilon$$
And if we stop before your last step, we get
$$\left |\dfrac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}-2 \right | \leq \frac{1}{\sqrt{n}}+\frac{2}{n}$$
Then, pick $N$. one choice might be $N=\lceil 4/\varepsilon^2\rceil$, in which case (for all $\varepsilon <1$)
$$\left |\dfrac{\sqrt{N}(1 + 8\sqrt{N})}{4n - 1}-2 \right | \leq \frac{1}{\sqrt{N}}+\frac{2}{N} \leq \frac{\varepsilon}{2}+\frac{\varepsilon^2}{2}< \varepsilon$$
And since $1/\sqrt{n}+2/n$ is decreasing in $n$, for any fixed $\varepsilon>0$, we have chosen an $N$ such that for all $n>N$
$$\left |\dfrac{\sqrt{N}(1 + 8\sqrt{N})}{4n - 1}-2 \right | < \varepsilon$$
And that is precisely the definition of a convergent sequence.
A: What you could have also done is
$$\left(\frac{\sqrt{x}(1 + 8\sqrt{x})}{4x - 1}\right)'=-\frac{4 \left(x+4 \sqrt{x}\right)+1}{2 (4 x-1)^2 \sqrt{x}} <0$$ So, the squence is decreasing.
Now, for large values of $n$
$$\frac{\sqrt{n}(1 + 8\sqrt{n})}{4n - 1}=2+\frac 1 {4\sqrt n}+O\left(\frac{1}{n}\right)$$
A: Don't be afraid to simplify even more, eliminate all additions if possible.
Get also rid of fractions by bounding them with an integer (here $\frac 23<1$).
For $n$ large then $2<\sqrt{n}$ (i.e. $n>4$) and $-1>-n$
$$\left|\frac{\sqrt{n}+2}{4n-1}\right|<\left|\frac{\sqrt{n}+\sqrt{n}}{4n-n}\right|=\left|\frac{2}{3\sqrt{n}}\right|<\frac 1{\sqrt{n}}\to 0$$ much easier to pick $N,\epsilon$ now.
