Evaluating the limit of $\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$ with definition I have troubles with "estimating".
We want to evaluate the limit with the definition.  For all $\epsilon >0$ we have to find an $N$, such that $|a_n-a|<\epsilon$ holds for $n>N$.
The limit in question is:
$$\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$$
We know the limit is $2$.

My approach:
Let $\epsilon > 0$.
$$\begin{align*} 
|a_n-2| 
&= \left|\frac{4n^2+9}{2n^2+2n+3} -2\right| \\
&= \left| \frac{4n^2+9 - 2(2n^2+2n+3)}{2n^2+2n+3} \right| \\
&= \left| \frac{4n^2+9 - 4n^2 - 4n -6}{2n^2+2n+3} \right| \\
&= \left| \frac{-4n+3}{2n^2+2n+3} \right| \\
&\leq \left| \frac{-4n}{2n^2+2n+3}\right|  \\
&\leq \frac{4n}{2n^2}  \\
&= \frac{2}{n} < \epsilon
\end{align*}$$
If we choose $N = \frac{2}{\epsilon}$ it holds that $\forall \epsilon > 0: |a_n -a| < \epsilon$, if $n > N$.

My professors approach:
$$\begin{align*}
\left| \frac{-4n+3}{2n^2+2n+3} \right| 
&\leq \left|\frac{3-4n}{2n^2} \right| \\
&\leq \left| \frac{3}{2n^2} \right| + \left|\frac{4n}{2n^2}\right| \\
&= \frac{3}{2} \cdot \frac{1}{n^2} + 2 \cdot \frac{1}{n}
\end{align*}$$
We note that:
$$\begin{align*}
&\frac{3}{2} \cdot \frac{1}{n^2} < \frac{\epsilon}{2} \Leftrightarrow \frac{3}{n^2} < \epsilon \iff n > \sqrt{\frac{1}{\epsilon}} \\
&2 \cdot \frac{1}{n}  < \frac{\epsilon}{2} \Leftrightarrow 4 \cdot \frac{1}{n} < 2 \iff \frac{4}{\epsilon} < n
\end{align*}$$
Then:
$$\begin{align*}
&\text{Let } N > \max\bigg\{\sqrt{\frac{1}{\epsilon}}, \frac{4}{\epsilon} \bigg\}  \\
&\implies\forall n > N \text{ it holds that } |a_n-a| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
\end{align*}$$

I have the following questions:
$1.$ What is wrong with my estimation?
$2.$ How does my prof get to $\frac{3}{2} \cdot \frac{1}{n^2} < \frac{\epsilon}{2}$
$3.$ What exactly does $\max\bigg\{\sqrt{\frac{1}{\epsilon}}, \frac{4}{\epsilon} \bigg\}$ tell us?

In your approach you need to find an upper bound for the numerator and a lower bound for the denominator.
Your calculation was ok but I would present it as below to make it a little bit more explicit:

*

*$4n-3<4n$

*$2n^2+2n+3>2n^2$
In fact your way of proceeding is totally fine, you can even skip the epsilon part and just conclude by $\frac 2n\to 0$.
In analysis except in rare exceptions when you need a really tight bound, don't bother with heterogeneous things like $\frac 32n^2+2\frac 1n$.
Just reduce it to a rough bound, i.e. $\frac 32<2$ and $\frac 1{n^2}<\frac 1n$ so that in the end you get $<\frac 4n$, and only then proceed with the epsilon if necessary.
I really don't get why professors go on with square roots of epsilons and so on...
So my advice, is just continue doing it the way you did !

Remark: in the same vein, when you have an epsilon-delta proof in $0$ and not infinity, you can also use this fact.
$x\to x_0$ then use $u=x-x_0\to 0$ in particular $u^n<\cdots<u^3<u^2<u<1$.
So you can bound any polynomial in $u$ by the sum of coefficients, i.e.
$|7u^3+2u^2-5u|<(7+2+5)|u|=14|u|\to 0$
Rather than bothering with $\epsilon, \sqrt{\epsilon}, \sqrt[3]{\epsilon}$...
