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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 ?

Michael Hardy
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optimal control
  • 535
  • 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. ${}\qquad{}$ – Michael Hardy Sep 11 '15 at 14:47

1 Answers1

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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 $0$.

hmakholm left over Monica
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