# Proving $\lim \limits_{x \to 2^-} \frac{x^2-7x+3}{x^2-4}$ using Delta-Epsilon

$$\lim \limits_{x \to 2^-} \frac{x^2-7x+3}{x^2-4}$$

The limit is $\infty$.

Trying to prove this via delta-epsilon has proven to be quite difficult.

The answer provided for this question goes like this: (of course translated to English)

Given Solution

"We want to prove that there exists $\delta \gt 0$ such that for every $x \in (2-\delta, 2)$ occurs $f(x)\gt M$.

We'll start by assuming $\delta \le 0.5$, and that $x^2-7x+3 \lt 0$ for $x\in (2-\delta, 2)$.

If x is in that section, then the following occurs:

$$\frac{x^2-7x+3}{(x-2)(x+2)} = \frac{x^2-7x+3}{(x+2)}* \frac{1}{x-2} \gt \frac{1}{x-2}*\frac{x^2-7x+3}{4} \gt \frac{1}{x-2}\frac{2^2-7*1.5+3}{4} \gt \frac{1}{2(2-x)}> \frac{1}{2\delta}$$

Then it is sufficient to pick $\delta=min({0.5, \frac{1}{2M}})$, and we get $\frac{1}{2\delta}\gt M$ like we wanted.

End of given solution

My questions are:

1.) Why did we pick $\delta \le 0.5$? How do we know which delta to pick? Based off of what?

2.) If someone could explain the thought process/chain of events that goes on with the inequalities that would be great

1) The reason why we choose an explicit number for $\delta$ is because we want to find a lower bound for the term $x^2-7x+3$. At this point, any $\delta>0$ will do the job, but due to the term $x-2$ on the denominator, we want to bound $x^2-7x+3$ by below a negative number (within the range $x\in (2-\delta,2)$) so that $\dfrac{1}{x-2}$ can becomes $\dfrac{1}{2-x}$.
2) But why all this work? The reason for this is because all we assume a-priori is that $2-x<\delta$ which is equivalent to $\dfrac{1}{2-x}>\dfrac{1}{\delta}$. Note that since we are evaluating left side limit, we have $2-x<\delta$ instead of the usual $|x-2|<\delta$.