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Let $g$ be a continuous function at a point $a \in \mathbb{R}$, and let

$$\lim_{x\rightarrow g(a)} f(x) = L$$

Show that

$$\lim_{x\rightarrow a} (f\circ g)(x) = \lim_{x\rightarrow g(a)} f(x)$$.

How do I prove this in the case that $f$ is not continuous. My current proof is incorrect since it assumes $f$ is continuous.

Let $\lim_{u\rightarrow g(a)} f(u) = L$. Also, since $g$ is continuous, $\lim_{x\rightarrow a} g(x) = g(a)$. If $\lim_{x\rightarrow a} (f\circ g)(x) = L$, then for all $\epsilon > 0$ there exists a $\delta > 0$ such that $0<|x - a|< \delta$ implies that $|(f\circ g)(x) - L | < \epsilon$.\

Now we will use the continuity of $g$ to prove the desired result. Since $\lim_{x\rightarrow a} g(x) = g(a)$, for all $\delta_1 > 0$ there exists a $\delta > 0$ such that $0<|x - a|< \delta$ implies that $|g(x) - g(a) | < \delta_1$.\

Since $\lim_{u\rightarrow g(a)} f(u) = L$, for all $\epsilon > 0$ there exists a $\delta_1 > 0$ such that $0<|u - g(a)|< \delta_1$ implies that $|f(u) - L | < \epsilon $.\

Now if we let $u = g(x)$, then we achieve the desired result since a $\delta$ and a $\delta_1$ can be chosen such that, $0 < |x - a| < \delta$ implies that $0 < |g(x) - g(a)|< \delta_1$, which finally implies that $|(f\circ g)(x) - L | < \epsilon$.\

Therefore $$ \lim_{x\rightarrow a} (f\circ g)(x) = \lim_{x \rightarrow g(a)} f(x). $$

user322087
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2 Answers2

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Always use the standard theorem for limit of composite functions:

Theorem: If $\lim\limits_{x \to a}g(x) = b$ and $g(x) \neq b$ in a certain deleted neighborhood of $a$ and $\lim\limits_{x \to b}f(x) = L$ then $$\lim_{x \to a}f(g(x)) = L$$

The result you have mentioned is false (for example when $g$ is a constant function say $g(x) = k$ and $f$ is not defined at $k$). A more complicated example of the failure of your result is when $g(0) = 0, g(x) = x\sin(1/x), x\neq 0$ and $f(x) = \dfrac{\sin x}{x}$ and $a = 0$.

Note that the above theorem is very powerful and is the basis of all substitutions used in evaluation of limits (for example replacing $x \to a$ with $x = a + h$ and $h \to 0$).

Also see this answer and related discussion.

Community
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Paramanand Singh
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  • "In a certain neighborhood of $a$"... You mean, "there exists at least one neighborhood of $a$ such that..." Right? The theorem appears to say that $\lim\limits_{x \to a}f(g(x))= \lim\limits_{y \to b} f(y)$ but not $\lim\limits_{x \to a}f(g(x))=f (\lim\limits_{x \to a}g(x))$, which is the usual theorem that's taught and which only works when $f$ is continuous on $b$. – William Jan 19 '22 at 11:22
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    @William: yes you have got the right meaning of that phrase. – Paramanand Singh Jan 19 '22 at 13:11
  • Paramand: It seems to me that the composition of limits hold under either of two different conditions: 1. The inner function $g(x) \neq b$ in a certain deleted neighborhood of $a$ (satisfied e.g. if $g$ has an inverse) OR 2: The outer function is continuous. Therefore, the OP's hypothesis that the outer function is continuous is sometimes necessary (e.g. for constant inner function). Do you agree? – SRobertJames Nov 25 '22 at 03:24
  • @SRobertJames: yes I agree. However it is better to explicitly state both the conditions. Without this using the word "necessary" may imply that continuity is necessary in all cases. – Paramanand Singh Nov 25 '22 at 04:33
  • @ParamanandSingh If limit of $f$ at $b$ and limit of $g$ at $a$ exist then can we claim limit of composition $f.g$ exist? – Meet Patel May 11 '23 at 03:43
  • @MeetPatel: the condition that $g$ does not equal its limit is necessary and with this condition the limit of composition exists. – Paramanand Singh May 11 '23 at 04:31
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Let $\varepsilon>0$ Since, $$\lim_{x\rightarrow g(a)} f(x) = L$$

There exists $\delta>0$ such that for $y\in \Bbb R$ the following is true

$$\color{blue}{|y-g(a)|<\delta \implies |f(y)-L|< \varepsilon }\tag{I}$$ Whereas $g$ is continuous at $x=a$ Then, there exists $\eta>0$ such for $x\in\Bbb R$ the following is true.

$$|x-a|<\eta \implies |g(x)-g(a)|<\delta\color{red}{\overset{y=g(x) ~~\text{in}~(I)}{\Longrightarrow }} |f(g(x))-L|<\varepsilon $$

That is for arbitrarily fixed epsilon there exists $\eta>0$ such that, for $x\in \Bbb R,$ $$|x-a|<\eta \implies |f(g(x))-L|<\varepsilon $$

Thus $$\lim_{x\to 0} f\circ g(x) = L = \lim_{x\to g(a)}f(x)$$

Guy Fsone
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