# Limit of a Composite Function

Definition of a limit I'm using:

Definition: Let $a$ be a limit point in the domain of a function $f$

Limit of the function f at a point $a\in\mathbb{R}\cup\{+\infty,-\infty\}$ equals $A\in\mathbb{R}\cup\{+\infty,-\infty\}$ iff $\lim{f(x_n)}=A$ for every sequence $\{x_n\}\subset D(f)\setminus\{a\}$ converging to $a$.

I am generally baffled by the theorem dealing with the limit of a composite function and its corollary.

Theorem: Let $f, g$ be functions, $a\in D(f\circ g)$ a limit point. If $\lim_{x\to a}g(x)=A$ and $\lim_{y\to A}f(y)=\alpha$, then $$\lim_{x\to a}(f\circ g)(x)=\alpha$$ Under the assumption that $g(x)\neq A$ for all $x\neq a$ in some neighbourhood of point $a$.

Question 1: Just to check if I'm understanding this correctly - the assumption here is needed because of the definition of a limit, where $A$ cannot be included in the sequences for the limit $\lim_{y\to A}f(y)$, right?

Corollary: Let $a\in D(f\circ g)$ be a limit point and $\lim_{x\to a}g(x)=A$. Then $$\lim_{x\to a}(f\circ g)(x)=\lim_{y\to A}f(y)$$ if at least one of these is true:

1. $f$ is continuous at $A$

2. $A=\pm\infty$

3. $g$ is strictly monotonic on some neighbourhood of point $a$

Question 2: This is where I am mainly confused. How do the three conditions arise from the theorem?

Thanks for any help!

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The assumption that $g(x)\neq A$ for $x$ in a neighborhood is needed. Take the domain $[0,1]$ and $f(0)=0$, and for $x>0$, $f(x)=1$. Take $g(x)=0$. Clearly $\lim_{y\rightarrow 0}f(x)=1$ and $\lim_{x\rightarrow 0}g(x)=0$, but $\lim_{x\rightarrow 0} (f\circ g)(x)=0\neq 1$.
For $f$ continuous, $\lim_{n\rightarrow\infty}f(x_n)=f(\lim_{n\rightarrow\infty}x_n)$. – Alex R. Jan 9 '13 at 20:03
Sorry for taking so long, I needed to review the topics before limits, as I felt my knowledge is not deep enough to understand this properly. Is the following then correct? (1) If $f$ is continuous, then we do not need to worry about $g(x)=A$ as $f(A)$ exists and is equal to the limits. (2) In this case, $g(x)$ can never equal to A $\forall x$. (3) If $g$ is strictly monotonous, then $g(x)=A$ only for one point and that point could only be $a$, which we are not considering, therefore $g(x)\neq A$ $\forall x\in D(f)\setminus\{a\}$ – Dahn Jahn Jan 19 '13 at 12:44