(This is an old question, but I still have some doubts on $\delta$)
To prove that $\lim_{x\to3} \frac{2x+3}{4x-9}=3$, I can find an appropriate $\delta$ ($\delta = \dfrac{\epsilon}{10}$) when $|4x-9|>1$, since:
$$ 10\dfrac{|x-3|}{|4x-9|}<\dfrac{\epsilon}{|4x-9|}< \epsilon.$$
However, I don't know how to proceed when $|4x-9|\le1$.
Note that for $x$ close to $2.25$ you get into trouble. To avoid this, the standard approach is to declare that no matter what, you will never pick $\delta$ to be larger than $\frac12$ (not that there is anything special with $\frac12$; any value below $\frac34$ will work). With this in mind, you only have to care about $x$ between $2.5$ and $3.5$.
On this interval we have $|4x-9|>1$, so $$ \left|\frac{2x+3}{4x-9}-3\right|=\left|\frac{-10x+30}{4x-9}\right|<10|x-3| $$ If we want the left-hand side to be less than some given $\varepsilon>0$, it's easy to see that it's enough to pick $\delta=\frac{\varepsilon}{10}$.
Thus our final choice for $\delta$ is $$ \delta=\min\left(\frac12, \frac{\varepsilon}{10}\right) $$