The full question:
Let $f(x) = \frac{P(x)}{Q(x)}$ be a rational function, where $P, Q$ are polynomials with no common factor of positive degree. Prove that $f$ has an infinite discontinuity at $x = a$, if and only if $Q(a) = 0.$
My attempt:
Intuitively, when $Q(a) = 0$, the vertical line $x = a$ is a vertical asymptote of $y = f(x)$. Thus $f(x)$ has an infinite discontinuity at $x = a$. One way this may be proven is to show that at least:
$\lim_{x \to a^+}f(x) = \infty \ \ or -\infty;$ or
$\lim_{x \to a^-}f(x) = \infty \ \ or -\infty;$
Let us consider the case of the right-sided infinite limit, where $P(x) > 0$ and $Q(x) > 0$ as $x \to a^+$ at all values of $x > a$ except at $x = a$ itself.
Then, to show $\lim_{x \to a^+}f(x) = \infty$, given a constant $K > 0$, we aim to find a $\delta > 0$ such that:
$$ 0 < |x - a| < \delta \Rightarrow f(x) = \frac{P(x)}{Q(x)} > K $$
To show that this only happens when $Q(a) = 0$, given $\epsilon > 0$, there exists a $\delta_1 > 0$ such that:
$$ 0 < |x - a| < \delta_1 \Rightarrow |Q(x)| < \epsilon $$
This may be worked out to be
$$ |Q(x)| < \epsilon \\ \frac{1}{|Q(x)|} > \frac{1}{\epsilon} $$
Since we take $P(x) > 0$ as $x \to a^+$, then,
$$ \frac{P(x)}{Q(x)} > \frac{1}{|Q(x)|} > \frac{1}{\epsilon} $$
Thus, we let $\delta = \delta_1$ and $K = \frac{1}{\epsilon}$ to show that $\lim_{x \to a^+} f(x) = \infty$ for $Q(a) = 0.$
I do not think that this flow of logic is correct since it involves making a few assumptions (i.e $Q(x)$ is continuous near $x = a$ and hence $\lim_{x \to a^+} Q(x) = Q(a) = 0$. Yet I am unsure of what direction to take to solve this question. I hope that I can obtain some tips from you guys. Thank you!
NB: I considered using proof by contradiction to establish the base statements, but I thought that by proving that $f$ does not have a limit at $x = a$ given $Q(a) = 0$, this does not necessarily establish that $f$ has an infinite discontinuity, since it may be a jump/'oscillating' discontinuity. Is this line of thinking correct as well?
Suppose that $Q(a)\ne0$. Then$$\lim_{x\to a}\frac{P(x)}{Q(x)}=\frac{P(a)}{Q(a)}\in\Bbb R$$and therefore there is no discontinuity at $a$.
Now, suppose that $Q(a)=0$. Then $(x-a)\mid Q(x)$. So, since $P(x)$ and $Q(x)$ have no common factors, $(x-a)\nmid P(x)$, and therefore $P(a)\ne0$. Take $K>0$. There is some $\delta_1>0$ such that$$|x-a|<\delta_1\implies\bigl|Q(x)\bigr|<\frac{|P(a)|}{2K}$$and there is a $\delta_2>0$ such that$$|x-a|<\delta_2\implies\bigl|P(x)\bigr|>\frac12\bigl|P(a)\bigr|.$$So, if $\delta=\min\{\delta_1,\delta_2\}$,$$|x-a|<\delta\implies\left|\frac{P(x)}{Q(x)}\right|>\frac{|P(a)|/2}{|P(a)|/(2K)}=K.$$Thus, I have proved that$$\lim_{x\to a}\left|\frac{P(x)}{Q(x)}\right|=\infty.$$
In your proof, I don't see where does the inequality$$\frac{P(x)}{Q(x)}>\frac1{|Q(x)|}$$come from.