Infinite Limit Theorem

Theorem
Let $f$ be a real function of $x$ of the form


 * $f \left({x}\right) = \dfrac {g \left({x}\right)} {h \left({x}\right)}$

Further, let $g$ and $h$ be continuous on some open interval $\mathbb I$, where $c$ is a constant in $\mathbb I$.

If:
 * $(1): \quad g \left({c}\right) \ne 0$
 * $(2): \quad h \left({c}\right) = 0$
 * $(3): \quad \forall x \in \mathbb I: x \ne c \implies h \left({x}\right) \ne 0$

then the limit of $f$ as $x \to c$ will not exist, and:


 * $\displaystyle \lim_{x \to c^+} f \left({x}\right) = +\infty$ or $-\infty$


 * $\displaystyle \lim_{x \to c^-} f \left({x}\right) = +\infty$ or $-\infty$

Proof
To prove the claim, it will suffice to show that for each $N \in \R_{>0}$, one can find an $\epsilon > 0$ such that:


 * $\left\vert{x - c}\right\vert < \epsilon \implies \left\vert{f \left({x}\right)}\right\vert \ge N$

So fix $N \in \R_{>0}$.

First, by continuity of $g$, find an $\epsilon_1 > 0$ such that:


 * $\left\vert{x - c}\right\vert < \epsilon_1 \implies \left\vert{g \left({x}\right) - g \left({c}\right)}\right\vert < \left\vert{\dfrac {g \left({c}\right)} 2}\right\vert$

which by Backwards Form of Triangle Inequality implies that:


 * $\left\vert{x - c}\right\vert < \epsilon_1 \implies \left\vert{g \left({c}\right)}\right\vert - \left\vert{g \left({x}\right)}\right\vert < \left\vert{\dfrac {g \left({c}\right)} 2}\right\vert$

and by rearranging, this becomes:


 * $\left\vert{x - c}\right\vert < \epsilon_1 \implies \left\vert{g \left({x}\right)}\right\vert > \left\vert{\dfrac {g \left({c}\right)} 2}\right\vert$

Next, by continuity of $h$, find an $\epsilon_2 > 0$ such that $\epsilon_1 > \epsilon_2$ and:


 * $\left\vert{x - c}\right\vert < \epsilon_2 \implies \left\vert{h \left({x}\right)}\right\vert < \dfrac {\left\vert{g \left({c}\right)}\right\vert} {2 N}$

From Reciprocal Function Strictly Decreasing, this means:


 * $\left\vert{x - c}\right\vert < \epsilon_2 \implies \dfrac 1 {\left\vert{h \left({x}\right)}\right\vert} > \dfrac {2 N} {\left\vert{g \left({c}\right)}\right\vert}$

These together imply that, whenever $\left\vert{x - c}\right\vert < \epsilon_2$, we have:

Thus $\epsilon_2$ is a valid choice for the sought $\epsilon$; hence the result.

Intuition
Consider the graph of the reciprocal function at the origin. Immediately to the right of the origin, for very small positive numbers, the function value gets very, very large. Similarly, immediately to the left of the origin, for very small (in magnitude) negative numbers, the function value gets very, very large (in magnitude), that is, very negative.

Note
This theorem is absolutely not saying that $\dfrac c 0 = \infty$ when dealing with real numbers. Division by zero is undefined.