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Let us consider two functions $f\left(t\right) = \frac{1}{t^2}$ and $g\left(t\right) = t^2$. We want to find the value of these two functions as $t\rightarrow \infty$. For the function $f$, this value is well defined and we write, $f\left(t\rightarrow\infty\right) = 0$. But for $g\left(t\right)$, the value lies at infinity. Which of the followings is true:

  1. The value of $g\left(t\rightarrow\infty\right)$ is not well defined.

  2. The value of $g\left(t\rightarrow\infty\right)$ lies at infinity.

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  • $\begingroup$ For the purposes of real analysis, it is common to introduce the extended real line, which includes both "endpoints" $+\infty$ and $-\infty$ in the sense of limiting values. These notions can be formally defined, in much the way that finite real numbers themselves are. $\endgroup$ – hardmath Sep 18 '14 at 18:38
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If we were working in the extended real numbers, one would typically continuously extend both of your formulae to set $f(+\infty) = 0$ and $g(+\infty) = +\infty$, rather than presuming that you meant the formula to apply only for real $x$.

Doing the same with the projective real numbers instead, you would do the same and have $f(\infty)=0$ and $g(\infty) = \infty$.


Regarding the notion of limit, while it is traditional to say in the case that

$$ \lim_{x \to +\infty} x^2 = +\infty$$

that the limit is undefined, in my opinion that is a rather incomplete treatment of the situation: if you're going to have a concept of a limit equaling $+\infty$, then you ought to do the thing properly and have $+\infty$ as an actual mathematical object and say that the limit is defined and equals $+\infty$.

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Usually, we say that a limit is undefined when it tends to $-\infty$ if you approach it from above/below and $+\infty$ when approaching it from the other side. If it tends to 'the same' $\infty$, we say the limit is well defined even if the object it is approaching doesn't belong to the real numbers. It might be slightly hand wavy but it's very convenient.

By the way, as a commenter hinted at, unless we actually extend the reals, you can't really talk about 'the value of $f(x \rightarrow \infty$)', since $f$ is only defined for real numbers, and obviously $\infty$ isn't part of the real numbers.

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  • $\begingroup$ But the cases posed in the Question involve $t \to \infty$, so the issue of approaching from two sides does not arise. $\endgroup$ – hardmath Sep 18 '14 at 18:40
  • $\begingroup$ It's not ambiguous 'which infinity' it's tending to in his case either, so the same reasoning applies $\endgroup$ – Benjamin Lindqvist Sep 18 '14 at 18:42
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As the term 'not well defined' you are using in 1) suggests, this is a matter of definitions. One common approach has been mentioned by hardmath, other approaches distinguish between $+\infty$ and $-\infty$, at least for the real number system.

It is, however, common (at least in analysis), to allow $\infty$ in one way or other as limit, and to this end provide a corresponding definition, like:

$$\lim\limits_{t \rightarrow a} f(t)=+\infty \, :\Leftrightarrow \forall N\in \mathbb{N}\,\exists \,\delta>0: |t-a| < \delta \Rightarrow f(t)>N $$.

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