Indeterminate form of infinity over 0?

Question

I know that indeterminate forms exist in limits, such as $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0^0$, $\infty^0$, $1^\infty$.

Then, if $\lim\limits_{x \to a} p(x)=\infty$ and $\lim\limits_{x \to a} f(x)=0$, can we call $\lim\limits_{x \to a} \frac{p(x)}{f(x)}$ an indeterminate form of type $\frac{\infty}{0}$? Or does it not exist since it has $0$ for the denominator?

Answer 1

This is not an indeterminate form, because it's clear what happens. If $f(x)$ approaches $0$ from above, then the limit of $\frac{p(x)}{f(x)}$ is infinity. If $f(x)$ approaches $0$ from below, then the limit of $\frac{p(x)}{f(x)}$ is negative infinity. If $f(x)$ keeps switching signs as it approaches zero, then the limit of the quotient fails to exist.

There's no "tug-of-war" here, like you have with indeterminate forms.

Answer 2

Yeah this is not an indeterminate form but it can be converted to $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

If $\lim\limits_{x \to a} p(x)=\infty$ and $\lim\limits_{x \to a} f(x)=0$, then $\lim\limits_{x \to a} \frac{p(x)}{f(x)}$ can be converted to indeterminate form via $\lim\limits_{x \to a} \frac{p(x)}{\frac{1}{f(x)}}$ (or) $\lim\limits_{x \to a} \frac{\frac{1}{p(x)}}{f(x)}$.