Composição de limites

In Continuity of Compositions, we saw that $f(g(x))$ is continuous at $p$ if $g(x)$ is continuous at $p$ and $f(x)$ is continuous at $g(p)$.

Let's study the corresponding statement about limits of compositions.

If $f(x),g(x)$ are functions with both limits $\lim_{x\to p} g(x) = L$ and $\lim_{x\to L} f(x) = M$ defined (so $L,M$ are real numbers), what can we say about the limit of their composition $f\circ g = f(g(x))$ at $p$, i.e. $\lim_{x\to p} f(g(x))$?

... or perhaps more intuitively,

If $f(x),g(x)$ are functions with $g(x)\to L$ as $x\to p$ and $f(x)\to M$ as $x\to L$ (here $L,M$ are real numbers), what can we say about $f\circ g = f(g(x))$ as $x\to p$?

How to read this post

Most of this post walks through intuition and some counterexamples. If you just want the core result of this post (which will be directly useful later on), you can skip ahead to the section Bright side: saved if $f(L) = M$ below.

Wait, isn't it the same as for compositions?

As $x\to p$, we have $g(x)\to L$, and as $y\to L$, we have $f(y)\to M$. So if we start with $x\to p$, then "plugging in" $y = g(x)$ should give us $f(g(x))\to M$, right?

No!

Where did our reasoning go wrong?

Let's look at this more closely (the following discussion will tell us how to find a counterexample too, not just the flaw in our reasoning!): if we have $$f(y)\to M \text{ as } y\to L,$$ why can't we "substitute" $y = g(x)$ to say $$f(g(x))\to M \text{ as } g(x) \to L \text{ (as } x\to p) ?$$ Well, remember that $\to$ (``tends to/approaches'') is just a sometimes convenient piece of notation. And here, it turned out to be more confusing than helpful.

But why? Some philosophy

But it works a little better for continuity than for limits, because when we say "$f(y)\to M$ as $y\to L$" in the context of limits, we have to ignore what happens at $y=L$: $f$ doesn't have to be defined at $y=L$, and even if $f(L)$ is defined, it doesn't have to equal $M$ (if it did, then $f$ would be continuous at $M$---which is basically why this all works better for continuity).

Finding and correcting the mistake in our flawed reasoning

So in our case, here's what's happening: if we have $$f(y)\to M \text{ as } y\to L, \text{ for $y\ne L$,}$$ we can only "substitute" $y = g(x)$ provided that $g(x)\ne L$: $$f(g(x))\to M \text{ as } g(x) \to L \text{ is valid for those $x$ with $g(x) \ne L$.}$$ (As discussed later, it can be valid for $x$ with $g(x) = L$ as well, basically when $f(L) = M$. Note that this is the same as saying $f$ is continuous at $L$.)

A counterexample to $f(g(x))\to M$, inspired by our flaw

This suggests that our counterexample should have some $x\approx p$ with $g(x) = L$. See if you can come up with one yourself!

Well, let's look at some $t\approx p$ with $g(t) = L$. Then intuitively, to get a counterexample, we should have $f(g(t)) = f(L)$ "far" from $M$. But that's relative to how close $t$ is to $p$. So to simplify things, let's just go ahead and take $g(x) = L$ for all $x$, and then take $f$ with $f(L) = M+9001$ (&). Then $f(g(x)) = f(L) = M+9001 \ne M$ for all $x$, certainly including (all!) $x$ close to $p$. So $f(g(x))\to M+9001 \ne M$ as $x\to p$, giving us a counterexample.

(&) There's one missing detail: we have to make sure this $f$ actually exists. But that's easy: to make sure $\lim_{x\to L} f(x) = M$, remember that the limit doesn't care about $x=L$. So we can just take $f(x) = M$ for $x\ne L$, which combined with $f(L) = M+9001$ completely defines $f$.

Bright side: saved if $f(L) = M$

But on the bright side, the fact that $\to$ is nicer under continuity tells us that we should consider the special case of continuous $f$ (continuous at $L$), where the "substitution" is actually valid. More precisely, we have the following very important result, which forms the basis for substituting to evaluate limits.

If $f,g$ are functions with $\lim_{x\to p} g(x) = L$ and $f(x)$ is continuous at $L$ (so $f(L) = \lim_{x\to L} f(x) = M$ in the notation above), then $\lim_{x\to p} f(g(x)) = f(L) = M$.

Do you see why? (How can we apply the correct reasoning from the earlier section Finding and correcting the mistake in our flawed reasoning?)

Since we have the annoying $y\ne L$ and $g(x)\ne L$ restrictions, let's forget about $\to$ notation for now, and just use $\approx$ instead. Then we know that

1. $g(x) \approx L$ for $x \ne p$ but close to $p$.

2. $f(y) \approx M$ for $y\ne L$ but close to $L$.

3. From 2, $f(g(x)) \approx M$ for $g(x) \ne L$ but close to $L$. By 1, this is guaranteed to happen whenever $x \ne p$ but is close to $p$, and also $g(x) \ne L$.

4. Wait, but if $g(x) = L$, then we automatically have $f(g(x)) = f(L) = M \approx M$. In particular, this holds for all $x\ne p$ close to $p$, but with $g(x) = L$ (complementing 3).

5. Combining 3 and 4, we're done! If $x\ne p$ is close to $p$, then whether or not $g(x) = L$, we still have $f(g(x)) \approx M$, as desired.

6. If you want, you can formalize this argument (and merge 3 and 4 into a single case) in the same way done here for Continuity of Compositions. Here's a sketch: $f(g(x))$ is close$_1$ to $M = f(L)$ whenever $g(x)$ is close$_2$ to $L$ (&&), and that's guaranteed to happen whenever $x$ is close$_3$ to $p$ but not actually equal to $p$. This is precisely what we wanted to prove!

(&&) Again, it doesn't matter if $g(x) = L$ or $g(x) \ne L$: that's the nice thing about continuity.

Can we classify all the cases where $f(g(x))\to M$?

Yes, and actually, we already covered all the intuition earlier, in the section Finding and correcting the mistake in our flawed reasoning. The idea is that our counterexample above doesn't need $g(t)=L$ for all $t$, but only "enough" $t$ close to $p$. Here "enough" means that for any small number $\alpha>0$ you pick, we need to be able to find a $t$ no farther than $\alpha$ away from $p$ (such that $g(t) = L$).

It turns out that this is the "most general" kind of counterexample that can occur. If you're interested, you can find the full details here, by searching "Theorem 13". But the most important part is that you understand the intuition we covered in this post.

I hope you found this insightful!

(Video) Limits of Composite Functions

by Tom Kaster

Here's a good video by Tom Kaster explaining how the limits of compositions work.