# Extending “as x approaches a” to “as g(x) approaches a”

All the definitions I can find of a limit (with functions from R to R) define something like:

"as x approaches a, f(x) approaches L"

Where x is treated as a variable that is quantified over in the definition.

Whereas many of these books then go on to use expressions of the form:

"as g(x) approaches a, f(x) approaches L"

without generalizing the definition appropriately.

Two questions:

1. what on earth makes this seem unproblematic to the authors? I'm guessing that the way I view things makes this use of notation seem more problematic than it is.

2. What is the appropriate formal defintion of the limit of f(x) as g(x) approaches a, where f:S->R and S is a subset of R.

-
If $g$ is invertible in a neighborhood of $a$ then there are no problems here since $f$ is locally defined as a function of $g$. To be honest, though, I don't know any examples where it isn't already clear that $f$ can be written as a function of $g$; can you clarify? – Qiaochu Yuan Mar 15 '11 at 2:04
Are you saying "lim f(x) as g(x)->a" is short for "lim f(g(x)) as x -> g^-1(a)"? If so, why does g need to be invertible in a neighborhod, rather than just at the point? (I mean, the second limit may not exist, but shouldn't that just tell us, then, that the first limit does not exist either?) – Quine42 Mar 15 '11 at 5:51

## 2 Answers

The delta-epsilon definition of the limit generalizes in a straightforward way. Suppose that for each $\epsilon > 0$, there exists $\delta(\epsilon) > 0$ such that $|g(x)-a|<\delta(\epsilon) \implies |f(x)-L|<\epsilon$. Then we say that $f(x)$ approaches $L$ as $g(x)$ approaches $a$. Though it might be more appropriate to say "whenever $g(x)$ approaches $a$", since this could happen for more than one value of $x$. For instance, $\sin(x)$ approaches $0$ whenever $\cos(x)$ approaches $1$.

-

I would interpret the statement that f(x) approaches L as g(x) approaches a to mean that for every $\epsilon > 0$ there exists $\delta > 0$ such that for all x, $0 < |g(x) - a| < \delta$ implies $|f(x) - L| < \epsilon$.

Note that x need not be a real variable in order to apply this definition, though doubtless examples of that type come to mind. For example, sin(x) tends to 0 as cos(x) approaches 1, but the converse is not true.

-
I'm not that surprised that someone gave the same answer as me, 12 seconds later, but the fact that you chose the exact same example is a little spooky :). – mjqxxxx Mar 15 '11 at 2:32
@mjqxxx: You were faster and had better LaTex! – hardmath Mar 15 '11 at 2:46
Wait, why is the converse not true? – Quine42 Mar 15 '11 at 3:20
Oh, and what's the domain of the implicit universal quantifier over x--am I correct that, in the first case, we are saying 'for all x in the domain of f', and in the second case, 'for all x in the domains of both f and g'? (Otherwise, a may be a limit point of domain(g) where every nbhd still contains points not in domain(g), which would make the limit undefined, even though it may be approaching a perfectly sensible limit--yes?) – Quine42 Mar 15 '11 at 6:16
@Quine42: Nice nym. The converse is not true because as sin(x) approaches 0, you get values of cos(x) near both &plusmn;1. I do assume f,g have a common domain. If not, then I agree an author owes you more of an explanation about what is meant by this phrase. – hardmath Mar 15 '11 at 13:19