Proof for The Limit $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4},\space n\in\Bbb{N}$ I am revising my knowledge on the topic of real analysis by attempting some simple proofs. The question requires for the proof of the Limit for $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4}$ using the definition of convergence: $$\forall\epsilon>0\space\space\exists N\in\Bbb{R}\space s.t\space\space \forall n>N,|s_n-L|<\epsilon$$ Below is my working: $$|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\epsilon$$ $$|\frac{3n^2-n}{4n^2+1}-\frac{3n^2+\frac{3}{4}}{4n^2+1}|=|\frac{-n-\frac{3}{4}}{4n^2+1}|=\frac{|-n-\frac{3}{4}|}{4n^2+1}<\frac{4n}{4n^2}=\frac{1}{n}<\epsilon $$ $$\frac{1}{n}<\epsilon\rightarrow\frac{1}{\epsilon}<n$$ $$\forall n>N=\frac{1}{\epsilon},\space|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}|<\frac{1}{n}<\epsilon$$ By the definition of convergence,$\space\frac{3n^2-n}{4n^2+1}\rightarrow\frac{3}{4}$ as $n\rightarrow\infty.$
Is this correct? Any feedback would be greatly appeciated.

Given $\epsilon>0$ be any number.
Now
\begin{align}
&\left|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}\right|\\
&=\left|\frac{-4n-3}{4(4n^2+1)}\right|\\
&=\frac{4n+3}{4(4n^2+1)}\\
&<\frac{n+1}{(4n^2+1)}\\
&<\frac{n+1}{(n^2+1)}\\
&<\frac{n+1}{n^2}<\epsilon\quad &\text{whenever}\quad n^2\epsilon-n-1>0\\
&&\text{i.e whenever}\quad n>\dfrac{1+\sqrt {1+4\epsilon}}{2\epsilon}\\
\end{align}
Now you can take $N=\left[{\dfrac{1+\sqrt {1+4\epsilon}}{2\epsilon}}\right]+1$ and this also allows $N$ to be a positive integer rather than just a real number which, in cases, can be difficult to find out.
Therefore
$\left|\frac{3n^2-n}{4n^2+1}-\frac{3}{4}\right|<\epsilon$ whenever $n>N$
