Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$ I am learning to prove limits with the epsilon-delta definition. This is officially the first exercise I complete on my own without peeking into the solution. My answer is different from the one in the solution - and to my horror, it is fairly larger than the one in the author's answer. But from what I gather, there is no single solution to these questions (I hope).
Could you take a look at my procedure? Is it correct?

Prove
$$\lim_{x\to3}\frac{x}{4x-9}=1$$

We want to prove that for some $\epsilon, \delta > 0$
$$|x-3| < \delta \iff \left| \frac{x}{4x-9}-1 \right| < \epsilon$$

Have
$$\left|\frac{x}{4x-9}-1 \right|< \epsilon$$
$$\implies \left|\frac{x-4x+9}{4x-9}\right|< \epsilon$$
$$\implies \left|(-3x+9) \cdot \frac{1}{4x-9}\right|< \epsilon$$
$$\implies |-3|\cdot|(x-3)| \cdot \left|\frac{1}{4x-9}\right|< \epsilon$$

I would like to get rid of that $\left|\frac{1}{4x-9}\right|$. First, notice that it is undefined for $x = \frac{9}{4}$.
Now then, keeping that in mind, let
$$|(x-3)| < \delta$$
Let's add another constraint:
$$|(x-3)| < \delta < \frac{9}{4}$$
After all, we can't let $x$ to undefine that fraction. Anyway, the above implies that
$$-\frac{9}{4} < x-3 < \frac{9}{4}$$
I want this $x-3$ to become $\frac{1}{4x-9}$, so we first multiply by $4$:
$$-9 < 4x-12 < 9$$
And add $3$:
$$-6 < 4x-9 < 6$$
And invert...
$$-\frac{1}{6} < \frac{1}{4x-9} < \frac{1}{6}$$
Cool, now we can replace $\frac{1}{4x-9}$ and end up with
$$|-3|\cdot|(x-3)| \cdot \left|-\frac{1}{6}\right|< \epsilon$$
Solve for $|(x-3)|$:
$$3\cdot|(x-3)| \cdot \frac{1}{6}< \epsilon$$
$$|(x-3)|< 2\epsilon$$

Answer:
$$\delta = \min\left\{\frac{9}{4},2\epsilon\right\}$$

The answer given in the solution is
$$\delta = \min\left\{\frac{1}{4},\frac{2}{3}\epsilon\right\}$$

As you can see, to my horror, the "range" of values in my solution is drastically greater than the one in the author's solution. Hence I am concerned that my answer is probably wrong.
 A: We want to show that for every $\epsilon\gt 0$, there exists a $\delta\gt 0$ such that if $0\lt |x-3| \lt \delta$ then 
$$\left|\frac{x}{4x-9}-1\right|\lt \epsilon.\tag{1}$$
So suppose we are given $\epsilon\gt 0$. We show how to find a suitable $\delta$.
Note that
$$\left|\frac{x}{4x-9}-1\right|=3|x-3|\left|\frac{1}{4x-9}\right|.$$
We can exercise control over $3|x-3|$ by choosing $\delta$ suitably. However, we must make sure that $\left|\frac{1}{4x-9}\right|$ does not spoil things by getting too large. So we want to make sure that $4x-9$ stays well away from $0$. 
Suppose for example that we insist that $\delta\le \frac{1}{4}$. If $|x-3|\lt \delta$, then $\frac{11}{4}\lt x\lt \frac{13}{4}$, and therefore $2\lt 4x-9\lt 3$.
It follows that if $\delta\le \frac{1}{4}$ then $\left|\frac{1}{4x-9}\right|\lt \frac{1}{2}$. Thus if $\delta\le \frac{1}{4}$ then 
$$ 3|x-3|\left|\frac{1}{4x-9}\right|\lt \frac{3}{2}|x-3|.$$
It follows that if $\delta=\min(\frac{1}{4}, \frac{2\epsilon}{3})$ then the desired inequality (1) holds. 
Remark: First of all, make sure that you use the correct definition of limit. Yours was not. Then write down the expression that you want to make "small" (less in absolute value than $\epsilon$.) In the formal write-up, there is no good reason to mention $\epsilon$ at that stage. Now we concentrate on the item that could cause difficulty. But if we choose $\delta$ small enough, we can check that it doesn't. I used $\frac{1}{4}$ because it makes the arithmetic simpler, but something like $\frac{1}{10}$ would work too. Verify that when $|x-3|$ is small (smaller than $\frac{1}{4}$, or smaller than $\frac{1}{10}$, the term $\frac{1}{4x-9}$ stays reasonable. Then we are close to finding a $\delta$ that will do the job.   
