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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 ?

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  • $\begingroup$ I changed every instance of \underset{x\rightarrow\infty}{lim} to \lim_{x\rightarrow\infty}. With the first you see $\displaystyle \underset{x\rightarrow\infty}{lim}$ and with the second you see $\displaystyle\lim_{x\rightarrow\infty}$. The latter is standard usage. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 11 '15 at 14:47
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$\frac{0}{\infty}$ 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 $0$.

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  • $\begingroup$ Is it the case also for the product of two functions that both have limit $\infty$ has limit $\infty$? $\endgroup$ – optimal control Sep 11 '15 at 15:15
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    $\begingroup$ @optimalcontrol: Yes, and you should be able to prove that directly from the definition of limit. $\endgroup$ – Henning Makholm Sep 11 '15 at 15:17
  • $\begingroup$ Wow this answer is just amazing! $\endgroup$ – Jinhua Wang May 17 at 10:31

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