Show $\lim_{x\to3} \frac{2x+3}{4x-9}=3$ by the definition of the limit. Show $\lim_{x\to3} \frac{2x+3}{4x-9}=3$ by the definition of the limit.
The definition of the limit: if for any given $\epsilon>0$ $\exists\delta>0$ such that if $x\in D(f(x))$ and $0<|x-c|<\delta$, then $|f(x)-L|<\epsilon$ where $L$ is the limit.
Following an example from the textbook, I have
$|f(x)-3|=\left|\frac{2x+3}{4x-9}-3\right|=\left|\frac{-6x+36}{3(4x-9)}\right|=\frac{6|-x+6|}{3|4x-9|}\cdot\frac{|x-3|}{|x-3|}=\frac{2|-x+6|}{|4x^2-21x+27|}\cdot|x-3|$
Restrict $|x-3|$ to the bounds $1<x<4$.  Then $|f(x)-3|\leq\frac{2|-(4)+6|}{|4(1)^2-21(1)+27|}\cdot|x-3|=\frac{2|2|}{|4-21+27|}\cdot|x-3|=\frac{4}{7}|x-3|$
Choose $\delta(\epsilon)=\inf\{1,\frac{7\epsilon}{4}\}$.  Then, is $0<|x-3|<\delta(\epsilon)$, we have $|f(x)-3|\leq\frac{4}{7}|x-3|<\epsilon$.
Thus, by the definition of the limit, $\lim_{x\to3} \frac{2x+3}{4x-9}=L=3$.
I don't understand what the point of all this was.  You can get the limit by essentially plugging in $x=3$. What exactly is being done? Why do we restrict $|x-3|$ to $1<x<4$? Why choose $\delta(\epsilon)=\inf\{1,\frac{7\epsilon}{4}\}$?
I get that if we end up with a form of $|f(x)-L|<\epsilon$, we then know $L$, but that's about all I grasped.
Any help would be greatly appreciated.

We have that
$$|f(x)-3|=\left|\frac{2x+3}{4x-9}-3\right|=\left|\frac{10(x-3)}{4x-9}\right|=\frac{10}{|4x-9|}|x-3|.$$
We want to simplify the denominator, ideally getting something like
$$|f(x)-3|<(...)|x-3|.$$
Restrict $x$ to a range around $3$ where $|4x-9|$ is bounded below, like $x\in(\frac{5}{2},\frac{7}{2})$. When $x$ is in that range, $|4x-9|>1$ so that
$$|f(x)-3|<10|x-3|.$$
Therefore if we are given $\epsilon>0$, and we want to provide a $\delta$ such that $|x-3|<\delta$ implies $|f(x)-3|$, we can take
$$\delta=\frac{\epsilon}{10}$$
so that $|f(x)-3|<10|x-3|<\epsilon.$
What about the restriction we placed on $x$? We've already done the hard work. When $\epsilon$ gets small, we have to shrink our interval $(3-\delta,3+\delta)$ so as to not let the function get too big there. We succeeded in quantifying that shrinking: $\delta=\epsilon/10$. When $\epsilon$ gets really big, we don't want to use that formula anymore: it gives us a big $\delta$ which may eventually include the asymptote at $x=9/4$. Fortunately, big $\epsilon$ is satisfied by the $\delta$ from any smaller $\epsilon$. Then we just cut off: if the formula would predict a $\delta>1/2$, don't even bother. Take
$$\delta=\textrm{min}\left(\frac{1}{2},\frac{\epsilon}{10}\right).$$
