I have a case where
$$\lim_{x\rightarrow\infty}=\frac{f\left(x\right)}{h\left(x\right)}$$
I know that $\lim_{x\rightarrow\infty} f(x)=0$ and $\lim_{x\rightarrow\infty} h(x)=\infty$
So at the and I have $\frac{0}{\infty}$.
I know that infinity is not a real number but I am not sure if the limit is indeterminate. (Also, there are people who are saying contradictory things on internet)
I know very well that it is not possible to use Hopital's rule.
My guess is that : As we know that $\lim_{x\rightarrow\infty}\frac{1}{\infty}=0$, We can just write $\lim_{x\rightarrow\infty} \frac{1}{\infty}=0$
$$\lim_{x\rightarrow\infty} \frac{1-0}{\infty}=0$$
$$\lim_{x\rightarrow\infty} \frac{1}{\infty}-\frac{0}{\infty}=0$$
So, in this case $\frac{0}{\infty}=0$.
What could be the answer and its explanation ?
$\endgroup$ 11 Answer
$\begingroup$$\frac{0}{\infty}$ is not an indeterminate form. On the contrary, those limits tell you that the limit of the entire quotient is $0$. This may be easier to see if you rewrite to $$ \lim_{x\to\infty} f(x)\frac1{h(x)} $$ where $\lim_{x\to\infty} f(x) = 0 $ and $\lim_{x\to\infty} \frac1{h(x)}=0 $, and the product of two functions that both have limit $0$ surely also has limit $0$.
$\endgroup$ 3
