# Why do some limits “explode” to infinity?

For example:

the limit of $e^t$ as $t$ approaches infinity is simply infinity.

but the limit of $e^{t^2}$ cannot be found because it explodes? Is this due to the fact that there are differing types of infinity and the second limit produces a non-"countable" infinity?

see referenced question

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The limit of e^(t^2) as t-->infinity is infinity, in the following sense: for any number N, if you choose t to be big enough, then e^(t^2) is more than N. This is exactly the same sense in which your first limit is infinity. –  Dan Ramras Oct 20 '10 at 1:38
On the other hand, one can talk about the asymptotics of a function, and $e^(t^2)$ is bigger than e^t in an asymptotic sense: if you consider their ratio $e^(t^2)/e^t = e^(t^2 - t)$, this still tends to infinity with t. –  Dan Ramras Oct 20 '10 at 1:41
I think you should read Noah Snyder's nice answer again in the question you referenced. There we are just counting/comparing the number of elements in sets. Here when we say the limit is infinity as $t$ approaches infinity, we mean that you cannot impede $f(t)$ to become large if you let $t$ gets bigger enough. –  Nuno Oct 20 '10 at 2:22
@Dan Ramras: If you write t^2 in curly brackets it works better: $e^{t^2}$. –  Hans Lundmark Oct 20 '10 at 7:06
Thanks, Hans! MathOverflow doesn't seem to have this problem... –  Dan Ramras Oct 20 '10 at 19:41

The limit can be found. What are we saying when we assert that $lim_{t\to\infty} f(t) = \infty$, with $f:\mathbb{R} \to \mathbb{R}$? What does it mean?

It means, by definition, that given $M >0$ there exists $\varepsilon >0$ such that if $t>\varepsilon$ then $f(t) > M$. Simply this. In (not so rigorous) words, for arbitrary large values of the parameter $t$ the function assumes arbitrary large values.

It is easy to see that the limit you want is indeed $\infty$ (in the sense just described). This can be seen noticing that $e^t < e^{t^2}$ for $t>1$.

I think that this question is not related to the linked one. For me, they are different things.

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To take a different perspective on it (if this is what you're looking for), there are limits that can't be found; the limit of $1/x$ as $x\rightarrow 0$ is one example. The limit doesn't exist because the limit coming from the right and from the left are different: one is $\infty$ and one is $-\infty$. I guess you could say that $1/x$ "explodes" at 0.

It's important to distinguish the type of infinity you're talking about, too. The "countable and uncountable" infinities you mention are cardinalities: they measure the number of elements in sets. The infinities you get with limits of functions like these are just limits of real numbers, and don't describe sets at all. You could call them "uncountable" since $\mathbb{R}$ is uncountable, I suppose, but then there wouldn't be a "countable" infinity in between $\infty$ and the real numbers. Basically, "infinity" can mean a lot of different things.

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Given your mentioning of the word "countable" (and looking at the link), I think that the "differing types of infinity" that you are thinking of are different infinite cardinalities (i.e. different possible infinite sizes of sets). This notion of infinity is not the same as the notion of infinity that is used when one says that a limit approaches infinity. (Of course, they are related at a basic level, in that both allude to the idea of being larger than any finite number, but at a more technical level, they are different notions.)

Ronaldo in his answer has given an explanation of what it means for $\displaystyle \lim_{t\to\infty} f(t)$ to equal $\infty$, and this is just as valid for $e^{t^2}$ as it is for $e^t$.

As Dan Ramras notes in one of his comments, the rate at which $e^{t^2}$ approaches infinity is much greater than the rate at which $e^t$ approaches infinty (which is turn much greater than the rate at which $t$ approaches infinity, for example). In terms of Ronaldo's answer, this means that for a given $M$, the $\epsilon$ that you have to choose to be sure that $e^{t^2} > M$ when $t > \epsilon$ is smaller than in the corresponding situation for $e^t$. But both quantities approach infinity as $t$ does.

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Some of your equations don't seem to have worked right. (Might be related to the doubled-dollar sign in the middle paragraph?) –  SamB Oct 20 '10 at 5:59
@SamB: Thanks for this. –  Matt E Oct 20 '10 at 14:58