$$\lim_{x\to2}\left(x^2+2x-7\right)\ = 1$$
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}
I think its more a matter of making language precise and remove any confusion. You are right that an informal meaning of $\lim_{x \to a}f(x) = L$ is that if the values of $x$ are near $a$ then values of $f(x)$ are near $L$.
How do we make this statement precise? We do that by quantifying the word "near". So the closeness of $x$ with $a$ is managed by number $\delta$ and closeness of $f(x)$ with $L$ is measured by $\epsilon$. We could have chosen any symbols say $A$ and $B$ instead of $\epsilon,\delta$ but choosing greek symbols gives you an air of uber-ness / geekiness / nerdiness. Thus mathematicians want to ensure that this concept is not to be taken lightly.
Now as we said earlier we want to ensure that when $x$ is close to $a$ then $f(x)$ should be close to $L$. This is like parents want that their kid studies hard to get good marks. The harder the kid studies the better the marks. Now it should be obvious to anyone that the goal here is "to get good marks" and "not just study hard".
So in case of limits the real goal is to ensure is that $f(x)$ gets close to $L$. The part of getting $x$ close to $a$ is only the means to an end. Therefore we have to give a bound $\epsilon$ for $|f(x) - L|$ and then determine $\delta$ such that $|x - a| < \delta$ would be sufficient to ensure the goal. When the goal fails (i.e. for some $\epsilon$ we are not able to get a corresponding $\delta$) we say that $L$ is not the limit of $f(x)$ as $x \to a$.
I should also point out the flaw with your argument. Suppose you challenge me with a $\delta$ and ask me to come up with an $\epsilon$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x - a| < \delta$. Then you have made my task so much easier and I will always win the challenge by choosing a very high value of $\epsilon$. Because you have given me complete leverage on the goal and I can choose to miss the goal by a wide margin (large value of $\epsilon$) whereas you keep on putting greater effort (choosing smaller $\delta$ for challenge). I hope you can understand this logic as to why I would win this challenge all the time if we play the game according to your rules.
Best Answer
Sure does. When $x$ is small and negative, the fraction is positive, so:
$$\frac{5x}{3x-9}<\epsilon$$ $$5x > (3x-9)\epsilon$$ $$5x > 3\epsilon x - 9\epsilon$$ $$(5-3\epsilon)x > -9\epsilon$$ $$x > \frac{-9\epsilon}{5-3\epsilon}$$
where the inequality sign was flipped because $3x-9$ must be negative under this assumption. The case for positive $x$ is much the same (and I'll leave it to you to do the details).