$$\lim_{x\to2}\left(x^2+2x-7\right)\ = 1$$
For every $\epsilon > 0$, there exists a $\delta >0$ such that
$|x-2| < \delta \implies |(x^2+2x-7) - 1| < \epsilon$.
Very often, you solve these problems by looking at what $\epsilon$ needs to do and then working backwards to what $\delta$ needs to do. In this case
$$|(x^2+2x-7) - 1| = |x^2+2x-8| = |(x+4)(x-2)| = |x+4|\,|x-2|$$
So, we need to make $|x+4|\,|x-2| < \epsilon$. We know we are going to make $|x-2| < \delta$, but what do we do with $|x+4|$?
\begin{align}
|x-2| < \delta
&\implies 2-\delta < x < 2 + \delta \\
&\implies 4-\delta < x + 4 < 4 + \delta
\end{align}
The trick is to limit the size of $\delta$. There is no fixed limit that you need to use. Just pick one. I think $10$ is a nice round number so I am going to say, suppose $0 < \delta < 2$. Then
\begin{align}
|x-2| < \delta \; \text{and} \; (0 < \delta < 2)
&\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0<\delta<2) \\
&\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (-2<-\delta<0)\\
&\implies 4 < x + 4 < 8 \\
&\implies |x+4| < 10 \\
&\implies |x+4||x-2| < 10\delta \\
\end{align}
You should see that we now solve $10\delta < \epsilon$ for $\delta$. We get $\delta < \dfrac{\epsilon}{10}$. But wait! We made an assumption that $\delta < 2$. That's very easy to fix. Our final formula is
$\delta = \min\left\{2, \dfrac{\epsilon}{10} \right\}$.
Then we get our proof by adding one more line to the previous argument.
\begin{align}
|x-2| < \delta \; \text{and} \; (0 < \delta < 2)
&\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0 < \delta < 2) \\
&\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (0<\delta<2) \\
&\implies 4 < x + 4 < 8 \\
&\implies |x+4| < 10 \\
&\implies |x+4||x-2| < 10\delta \\
&\implies |(x^2+2x-7) - 1| < \epsilon
\end{align}
Let $|x| \lt 1/5$, then
1)$-1/5< x < 1/5$, or
$-3/5 +1< 3x +1<3/5+1$;
2) $|2x+3| \le 2|x| +3\lt 17/5$.
Let $\epsilon>0.$
Choose $\delta = \min(1/5, (2/17)\epsilon )$, then
$|x|\lt \delta$ implies
$\dfrac{|2x+3|}{|3x+1|}|x| \lt\dfrac{17/5}{2/5}|x| =$
$(17/2)\delta \lt \epsilon$.
Best Answer
You have\begin{align}\left|\frac8{9x-3}\right|<\varepsilon&\iff|9x-3|>\frac8\varepsilon\\&\iff\left|x-\frac13\right|>\frac8{9\varepsilon}.\end{align}If $x\geqslant\dfrac13$, this is equivalent to $x>\dfrac8{9\varepsilon}+\dfrac13$. So, take $N=\max\left\{\dfrac13,\dfrac8{9\varepsilon}+\dfrac13\right\}$, and then$$x>N\implies\left|\frac{x-3}{3x-1}-\frac13\right|<\varepsilon.$$