Can one infinity be greater than other? [closed]

Both these limits tend to infinity but it is obvious to say that $$\lim_{x \to 0}\frac{2}{x} \gt \lim_{x \to 0}\frac{1}{x}$$
as at any point it is true, if not what about these one
$$\lim_{x \to 0}\frac{1}{x^2} \gt \lim_{x \to 0}\frac{1}{x}$$
$$\lim_{x \to 0}\frac{1}{x^3} \gt \lim_{x \to 0}\frac{1}{x}$$
If all infinities are equal then all graphs mus intersect at a point which is never true

closed as off-topic by Asaf Karagila♦, TheGeekGreek, Saad, ahulpke, SchrodingersCatMay 11 '18 at 14:55

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• what if $x$ convergence to zero from below (think of $x$ as a very small negative number, so the first and third inequality makes no sense) – Yanko May 11 '18 at 11:05
• Caution: the limits are equal for $x \to \infty$. The strict inequality holds only for any $x \not =0$ fixed. – YukiJ May 11 '18 at 11:06
• in this case it converges from $0^+$ side – Abhishek May 11 '18 at 11:06
• how can they be equal – Abhishek May 11 '18 at 11:07
• Possible duplicate of Is it faster to count to the infinite going one by one or two by two? – Asaf Karagila May 11 '18 at 11:18

As already noted $\infty$ is used in limits as a symbolic manner to express that the function becomes larger than any fixed bound $M$ as $x\to 0$ (or smaller for the case $-\infty$).

What we can to compare two different functions is to consider their ratio for a same value of $x$ and also take the limit of that ratio, that is for example

• $f(x)=\frac1{x^2}$
• $g(x)=\frac1{x^4}$

then

$$\lim_{x \to 0}\frac{f(x)}{g(x)}=\lim_{x \to 0}\frac{\frac1{x^2}}{\frac1{x^4}}=\lim_{x \to 0} \frac{x^4}{x^2}=\lim_{x \to 0} x^2=0$$

then we say that $g(x)$ tends to $\infty$ faster than $f(x)$ for $x \to 0$.

Depending on your concept of infinity, yes, one infinity can be greater than another. However, when taking limits of real-valued functions, there really is only one infinity. That infinity means "The function grows beyond any finite bound", and that's it. There is, in this respect, no notion of how fast it grows beyond any finite bound; they all reach the same $\infty$ in the end.

One case where infinities have different sizes are when they measure the size of sets. The infinity which describes the number of elements in the set of natural numbers is strictly smaller than the infinity which describes the number of elements in the set of the real numbers.

• I agree with you but if we consider infinity to be just a number then one can be greater than other – Abhishek May 11 '18 at 11:09
• @abhishekchaudhary But infinity isn't a number. Besides, does $\lim_{x\to 0}x^2$ reach a smaller $0$ than $\lim_{x\to 0}x$? It doesn't make sense to differentiate in the large end and not the small end. – Arthur May 11 '18 at 11:12
• @abhishekchaudhary That being said, I'm sure there are theories out there which deal with this in a different way, such that you do indeed get different infinities (hyperreal numbers might be one, I don't know). But that would be venturing into non-standard analysis, and I personally have little experience with that. – Arthur May 11 '18 at 11:14

Infinities can be compared by calculating $$\lim_{x \to a} \frac{f\left(x\right)}{x}, f\left(a\right) = \infty$$
if they are also $\infty$ then calculate $$\lim_{x \to a} \frac{f\left(x\right)}{x^2}, f\left(a\right) = \infty$$
and so on this can compare two infinities

• Did you ask a question just to post an answer? – Asaf Karagila May 11 '18 at 11:27
• i had answer i just wanted to know if anyone else has the same answer – Abhishek May 11 '18 at 11:34

No.

Infinity is a concept, not a number. Therefore, the limits, while one accelerates at a faster rate, both have the same endpoint of infinity, and therefore are equal.

Have a look at this for a clearer explanation of why $\infty$ is not a number.