General proof of limit composition theorem on continuous function

Question

Let $A, B \subset \mathbb{R}$, $a, b, c \in \overline{\mathbb{R}}$, $a$ and $b$ be limit points of $A$ and $B$. Let $f: A \rightarrow B$ and $g : B \rightarrow \mathbb{R}$. I have to prove that if $b \in B$ and $g$ is cointinous in $b$ and

$\lim \limits_{x \to a}{f(x)} = b$
$\lim \limits_{y \to b}{g(y)} = c$

then $\lim \limits_{x \to a}{(g \circ f)(x)} = c$. How to make a proof to as general theorem as above?

Hint: Show $\lim_{x\to a}f(x)=c\iff$ for any $(x_n)\to a$, $f(x_n)\to c$ — John, Jan 15 '15 at 16:57
You should see the answer http://math.stackexchange.com/a/1073047/72031 which shows that continuity of $g$ is not necessary. — Paramanand Singh, Jan 17 '15 at 05:43

idm · Answer 1 · 2015-01-15T17:06:45.750

3

Let $\varepsilon>0$.

By the second limit, $|g(y)-c|<\varepsilon$ for all $|y-b|<\delta$ for a certain $\delta>0$. By the first limit, $|f(x)-b|<\delta$ for all $|x-a|<\delta'$ for a certain $\delta'>0$. And thus, if $|x-a|<\delta'$, $$|f(x)-b|<\delta,$$ therefore $$|g(f(x))-c|<\varepsilon$$ if $|x-a|<\delta'$.

edited Jan 15 '15 at 17:06

answered Jan 15 '15 at 17:01

idm

11,824

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note that c could be $\infty$ – alex Jan 15 '15 at 17:08
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Do you think continuity of $g$ at $b$ is needed in this question? – John Jan 15 '15 at 17:09
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absolutely not ! The continuity is not necessary ! it would be important if $c=g(b)$, but not here. – idm Jan 15 '15 at 17:10
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Yes because $\lim \limits_{y \to b}{g(y)}$ don't need to be $g(b)$ otherwise. – alex Jan 15 '15 at 17:11
but $\lim_{y\to b}g(y)\neq g(b)$ a priori. – idm Jan 15 '15 at 17:12
If $g(y)$ is continous in $b$ then $\lim_{y\to b}g(y)= g(b)$ by definition. – alex Jan 15 '15 at 17:15
right but your hypothesis is $\lim_{y\to b}g(y)=c$ not $g(b)$ ! – idm Jan 15 '15 at 17:35
Yes, but the assumption is nessesary. – alex Jan 15 '15 at 17:40
no it's not necessary ! – idm Jan 15 '15 at 18:20
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@idm The continuity is necessary. Your proof fails because the definition of function limit is when $\mathbf{0<}|y-b|<\delta$ we have $|g(y)-c|<\epsilon$. Counterexample can be found here: https://math.stackexchange.com/questions/2069972/composition-of-limits – MathEric Mar 12 '20 at 23:59

General proof of limit composition theorem on continuous function

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