Note that I got this question from just messing around with a couple of functions and don't know if it is actually true.
Let $f$ be a function such that $\lim\limits_{x \to a}|x-a|f(x)=L$, and $0<L<\infty$. Additionally let $f$ be integrable in a region, $B$, where $a$ is an accumulation point of $B$. Then we know that, $$\forall \epsilon >0 \ \exists \delta >0 : 0<|x-a|<\delta \implies ||x-a|f(x)-L|<\epsilon.$$ Fix $\epsilon$ such that $L>\epsilon$, then this means that, \begin{equation} \frac{L-\epsilon}{|x-a|}<f(x)<\frac{L+\epsilon}{|x-a|}. \end{equation} Then let $F$, the anti-derivative of $f$ be defined as, $$F(x)=\int_{d}^{x}f(t) \ \mathrm{d}t + k,$$ for some $d \in B$ and $k \in \mathbb{R}$. Then I aim to show that $\lim\limits_{x \to a}F(x)=\infty$.
Now we know that $\dfrac{1}{|x-a|}$ is integrable everywhere except $x=a$. Additionally $$\int_{y}^{a}\dfrac{1}{|x-a|} \ \mathrm{d}x,$$ is unbounded and goes to $\infty$ for $y \neq a$. Then taking the left hand side of the inequality above and integrating from $d<a$ to $x$ where, where $x$ is in $B$, we get that, $$(L-\epsilon)\int_{d}^{x}\frac{1}{|t-a|} \ \mathrm{d}t+k<\int_{d}^{x}f(t) \ \mathrm{d}t+k=F(x).$$ Now taking the limit as $x$ approaches $a$ on both sides we achieve the desired result. That is $\lim\limits_{x \to a}F(x)=\infty$.
$\textbf{My question is:}$
Is this proof valid? Initially I thought that $L>1$ was required, but it seems this 'proof' doesn't use this assumption anywhere so I omitted it. It seems intuitively correct to assume this as the area under $\dfrac{1}{|x-a|}$ is unbounded if we take one of the bounds as $a$, so the area under $f(x)$ must be unbounded as well.
$\rule{8cm}{0.4pt}$ $\textbf{Removed but to provide clarity to one of the answers.}$
The function $P(x)=\dfrac{1}{|x-a|}$ was chosen somewhat arbitrarily, since it has that unboundedness behaviour for the area under the curve near $x=a$. Is there any function that has the same property but less than $P(x)$? Then perhaps $f(x)$ doesn't have as strict of a restriction as before, as we can plug in this new function for $P(x)$.
For (3) we have $\displaystyle\frac{1}{(t-a)|\log(t-a)|} < \frac{1}{t-a}$ for $a < t < a+e$, and for all $a < x < a+e$,
$$\int_a^x\frac{dt}{(t-a)|\log(t-a)|}= \int_a^x\frac{dt}{t-a} = +\infty$$
Also, just as there is no smallest divergent series, since $\sum \frac{1}{n }$, $\sum \frac{1}{n\log n}$, $\sum \frac{1}{n\log n\log \log n}$, etc. diverge, there is no smallest function for which the improper integral diverges.