16
$\begingroup$

After answering this question, I was wondering if the following generalization holds true:

Claim: If $\lim \limits_{x\to a}g(x)=b$, then $\lim \limits_{x\to a}f(g(x))=\lim \limits_{y\to b}f(y)$.

I've seen some people use this change of variables before when evaluating difficult limits, but I haven't seen this presented as a theorem in a textbook. Is this claim true? If not, is the claim salvageable with additional hypotheses? For example, must $f$ be continuous for the claim to hold?

$\endgroup$
2

2 Answers 2

13
$\begingroup$

You certainly need some assumptions on $f,g$, as Jared shows.

THM Suppose that $f(y)\to \ell$ as $ y\to b$. Suppose that ${\rm im}\, g\subseteq {\rm dom}\, f$, and suppose that $g(x)\to b$ as $x\to a$, yet $g$ does not attain the value $b$ in a neighborhood $B(x,\eta)-\{x\}$. Then $$f\circ g(x)\to \ell \;\;\text{ as } x\to a$$

P Let $\epsilon >0$ be given. Since $f\to\ell $ as $y\to b$ there exists $\delta >0$ such that $0<|y-b|<\delta$ implies $|f(y)-\ell|<\epsilon$. Since $g\to b$ as $x\to a$, there exists $\eta >\delta'>0$ such that $0<|x-a|<\delta'$ implies $0<|g(x)-b|<\delta$. But then we will have $|f(g(x))- \ell|<\epsilon$ whenever $0<|x-a|<\delta'$, so the claim follows. $\blacktriangle$

This then gives the standard

COR Suppose that $f$ is continuous at $g(a)$ and $g$ is continuous at $a$. Then $f\circ g$ is continuous at $a$.

$\endgroup$
6
  • $\begingroup$ (Note this answer was posted before Jared added "If we suppose...", so it was intended as a complement of his counterexample) $\endgroup$
    – Pedro
    Jul 31, 2013 at 0:46
  • 1
    $\begingroup$ The definition of $f\to\ell$ as $x\to a$ is that for all $\epsilon>0$, there exists $\delta>0$ such that $0<|x-a|<\delta \implies |f(x)-\ell|<\epsilon$. I've noticed that both you and Jared omitted the part where $0<|x-a|$. Is this part unimportant? I can see that: $$ 0<|x-a|<\delta' \implies |x-a|<\delta' \implies |g(x)-b|<\delta $$ but how do we do know that $g(x) \neq b$ so that we can use the fact that: $$ 0<|g(x)-b|<\delta \implies |f(g(x))-\ell|<\epsilon $$ $\endgroup$
    – Adriano
    Jul 31, 2013 at 1:05
  • 1
    $\begingroup$ @Adriano You're absolutely correct. Although it is certainly a minor detail. $\endgroup$
    – Pedro
    Jul 31, 2013 at 1:08
  • 1
    $\begingroup$ In the theorem statement, I think the neighborhood has to be of the point a, right? $\endgroup$ Jan 26, 2020 at 22:18
  • 1
    $\begingroup$ this is probably the most used and most unkwnon theorem $\endgroup$
    – frhack
    Jan 9 at 11:54
8
$\begingroup$

A counterexample. Let $a=b=0$, $g(x)=x^2$, and let

$$f(x)=\begin{cases}1&x\ge 0\\0&x<0\end{cases}$$

Then $\lim_{y\to 0}f(y)$ does not exist, but $\lim_{x\to 0}f(g(x))=1$.

If we suppose that $\lim_{y\to b}f(y)$ exists, then the result holds.

First suppose that the limit is finite and equal to $L$. Fix $\epsilon>0$, and let $\delta$ be such that

$$|y-b|<\delta\Longrightarrow |f(y)-L|<\epsilon$$

Now, choose $\delta'$ such that

$$|x-a|<\delta'\Longrightarrow|g(x)-b|<\delta$$

Notice now that we have

$$|x-a|<\delta'\Longrightarrow|f(g(x))-L|<\epsilon$$

which shows that $\lim_{x\to a}f(g(x))=L$. The case of an infinite limit is shown similarly.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.