Can one non-cardinal infinity be greater than other non-cardinal infinity? As far as I know, there are two different notions to the word "infinity" in Mathematics.
First notion of infinity has to do with the cardinality of a set: if a set contains infinite number of elements, the set is said to be an "infinite set"; this notion of infinity deals with "how many".
Second notion of infinity deals with "how large" rather than "how many" - here, infinity is simply a "infinitely large number", that is, one that is greater than any numbers.
I have heard that some mathematician proved that one infinity can be larger than other infinity, but I have also heard that the proof only has to do with the first notion of infinity.
What about in terms of the second notion of infinity - say you have two infinitely large numbers (again, this is different than having a set with infinite number of elements). Can one infinitely large number be greater than other infinitely large number? 
 A: Your second question needs clarification (what exactly do you mean by "number"?), but here's one interpretation:
An ordered field is a set $S$ together with operations $+$ and $\times$, and a binary relation $<$, such that


*

*$(S, +, \times)$ is a field

*$<$ is a total order on $S$, and

*$<$ is "compatible" with the field structure (e.g. $a<b$ implies $a+c<b+c$, etc.)
An ordered field is Archimedean if for each $x>0$ there is some natural number $n$ such that $1+1+...$ ($n$-many times) is $>x$. For example, $\mathbb{R}$ is an ordered field. Perhaps surprisingly, there are lots of non-Archimedean ordered fields! In such fields, an element $x$ is often called "infinite" if $x>1+1+...$ ($n$ times) for every natural number $n$. Now infinite elements can be distinguished - e.g. if $x$ is infinite, so is $x+1$, and $x<x+1$.

Note that this is not what is meant when people talk about limits at infinity - there, if we're even treating "$\infty$" as a real thing and not just a convenient shorthand, we're considering a number system (usually called the extended real line) $\mathbb{R}\cup\{\infty, -\infty\}$. Among other things, this system satisfies $\infty+1=\infty$. It's easy to check (and a good exercise) that this system is not a field.

Tl;dr: it depends what you mean by "number." Note that, by contrast, cardinality of a set is precisely defined.
A: I think you're confused about your "second type of infinity".    In talking about limits of or at infinity, infinity is not regarded as a number, rather having an infinite limit as, say, $x \to 0$ or $x \to +\infty$ is a way a function c\inftyan behave when $x$ is close to $0$ or very large.  
In that regard, there is a sense in which one "infinity" is greater than another, although it isn't a comparison of two "infinite numbers", rather comparison of the way two functions of $x$ behave in the limit.  You might look up Big O notation.
A: Assuming that you are referring to transfinite ordinals as the second category of infinity, that I should a good question, but that can be solved using diagonalization (of sorts), the set of natural numbers: ω, is (0), 1, 2, 3... and once you’ve ‘listed them all’, there is no more (even though it is infinite, for the sake of simplicity, imagine you are at the end. Now we can prove that the set of all real numbers contains more elements. Say you listed every single infinite decimal, so you found the ‘end’, but you can define a new infinite’s decimal, by taking the first digit of the first decimal and adding one (or if it was 9, subtract 1), and putting the result as the first digit of the new decimal, repeat this process for all ω decimals, and you’ll have a new one. So actually the set is ω + 1, but then repeat the entire process again, to make a new one. You could do this an infinite number of times, yet still be able to make new ones. Therefore there is more elements within the set of real numbers than her set of natural numbers, solving your question and proving that a non-cardinal infinity can be larger than another one.
I know you said that you weren’t talking about ordinal numbers, but both definitions were of cardinal numbers so I decided to go with the question in the title instead.
