"plugging in" infinity Do you allow your undergraduate calculus students to "plug in" infinity when evaluating a limit at infinity?
For example, would you accept the notation: $\lim\limits_{x\to\infty}\frac1x=\frac1\infty=0$?
There is disagreement among my colleagues.
I'm in the camp that does not accept this abuse of notation for three main reasons:


*

*Infinity is not a constant and therefore direct substitution does not apply.

*"Plugging in" infinity defeats the very notion of a limit because we no longer see the concept of the unbounded variable if the variable has been replaced.

*This may lead to further misunderstanding of the concept of a limit.

 A: 
I'm in the camp that does not accept this abuse of notation for three main reasons:

I don't really agree that this is an abuse of notation, but as far as I know, you only really get "half" the result you're looking for like this.
Let me explain.
Let $P$ denote the set obtained by adjoining a point called $\infty$ to $\mathbb{R}$. This can be made into a topological space in the usual way, thereby making $\mathbb{R}$ into a topological circle. Each partial function $f : \mathbb{R} \rightarrow \mathbb{R}$ defines a function $\hat{f} : P \rightarrow \mathcal{P}(P)$ as follows: $y \in \hat{f}(x)$ iff there exists a sequence $X$ in the domain of $f$ such that $X$ converges to $x$ in $P$ and $f(x)$ converges to $y$ in $P$. This can in turn be extended to a function $\hat{f} : \mathcal{P}(P) \rightarrow \mathcal{P}(P)$ by writing $\hat{f}(A) = \{\hat{f}(a) : a \in A\}.$
For example:
$$\frac{1}{0} = \infty, \qquad \frac{1}{\infty} = 0$$
Similar things can be done with multi-input partial functions $\mathbb{R}^n \rightarrow \mathbb{R}$, like division. We get equations like $$\frac{\infty}{\infty} = P.$$
With these definitions, we can safely reason as follows:
$$\lim_{x \rightarrow x_0} y \in (x:=x_0)y$$
for any $x_0 \in P$, where $(x:=x_0)$ means the result of replacing every copy of $x$ with $x_0$ i.e. uniform substitution.
For example, the following reasoning is perfectly valid
$$\lim_{x \rightarrow \infty} \frac{5+(1/x)}{x} \in \frac{5+(1/\infty)}{\infty} = \frac{5+0}{\infty} = \frac{5}{\infty} = 0$$
This tells us that if the limit exists, then it's zero. What it seems not to tell us is that the limit actually exists.
Thoughts/comments everyone?
