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 $limlimits_{x to a}|x-a|f(x)=L$, 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 $limlimits_{x to a}F(x)=infty$.

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 $limlimits_{x to a}F(x)=infty$.

$textbf{My first question is:}$

Is this proof valid? Initially I thought that $L>1$ 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.

$textbf{Second question (and the one I’m more concerned with):}$

Supposing the fact that this is true. Then couldn’t I supposedly assume the following proposition.

Let $g(x)$ and $f(x)$ have the following properties described in the proof, that is as $x$ approaches $a$, $G(x)$ and $F(x)$ (their respective anti-derivatives) are unbounded and approach $infty$.

Then if we take the limit of the ratio of $g(x)$ and $f(x)$, ($G(x), g(x) neq 0$),

$$lim_{x to a} frac{f(x)}{g(x)}.$$

Then integrate the numerator and denominator to obtain a supposedly different limit, that is,

$$lim_{x to a} frac{F(x)}{G(x)}.$$

However since we know that $F(x)$ and $G(x)$ both go to $infty$ as $x$ approaches $a$ we can apply L’Hospital’s rule and supposedly achieve the following,

$$lim_{x to a} frac{F(x)}{G(x)}=lim_{x to a} frac{f(x)}{g(x)}.$$

That is to say we achieve a sort of “inverse” L’Hopital’s rule if you will, assuming that the functions that are being concerned obey the properties laid out earlier.

$textbf{Third and final question:}$

The function $dfrac{1}{|x-a|}$ was chosen somewhat arbitrarily since it has that nature of having an unbounded area. Is there any function perhaps smaller than this but also has an unbounded area? Then perhaps this condition would capture a wider array of functions.

The reason for not being able to assume that solely $f(x)$ goes to infinity as $x$ goes to $a$ is because counter-examples exist for this, that is $ln|x|$, and a more strict unboundedness needed to be placed.