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?


2 Answers 2


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 $a$.

  • $\begingroup$ (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
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    $\begingroup$ 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
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    $\begingroup$ @Adriano You're absolutely correct. Although it is certainly a minor detail. $\endgroup$
    – Pedro
    Jul 31, 2013 at 1:08
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    $\begingroup$ 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

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


Notice now that we have


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


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