The continuity law for composition states, informally, that:
$$\text{IF } f \text{ is continuous at } a \ \text{ AND }\ g \text{ is continuous at } f(a) \text{ THEN } g(f(a)) \text{ is continuous at } a$$
Or, using the limit definition of continuity:
$$ \lim_{x \to a} f(x) = f(a) \wedge \lim_{x \to f(a)} g(x) = g(f(a)) \Longrightarrow \lim_{x \to a} g(f(x)) = g(f(a))$$
Or, using the formal epsilon-delta definitions:
$$ \forall \epsilon > 0, \exists \delta_1 > 0, \forall x \in \mathbb{R}, |x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \epsilon \tag{1}\label{1}$$
$$ \forall \epsilon > 0, \exists \delta_2 > 0, \forall x \in \mathbb{R}, |x – f(a)| < \delta_2 \Longrightarrow |g(x) – g(f(a))| < \epsilon \tag{2}\label{2}$$
$$ \forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathbb{R}, |x – a| < \delta \Longrightarrow |g(f(x)) – g(f(a))| < \epsilon \tag{3}\label{3}$$
$$ \eqref{1} \wedge \eqref{2} \Longrightarrow \eqref{3}$$
Proof:
We assume \eqref{1} and \eqref{2}. We want to show \eqref{3}.
Let $\epsilon > 0$.
We will use the same $\epsilon$ in \eqref{2}, giving us:
$$ \exists \delta_2 > 0, \forall x \in \mathbb{R}, |x – f(a)| < \delta_2 \Longrightarrow |g(x) – g(f(a))| < \epsilon \tag{2a}\label{2a} $$
Let $\delta_2 > 0$ such that the following is true:
$$ \forall x \in \mathbb{R}, |x – f(a)| < \delta_2 \Longrightarrow |g(x) – g(f(a))| < \epsilon \tag{2b}\label{2b}$$
We will substitute $\delta_2$ for $\epsilon$ in $\eqref{1}$ to obtain:
$$\exists \delta_1 > 0, \forall x \in \mathbb{R}, |x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \delta_2\tag{1a}\label{1a} $$
Let $\delta_1 > 0$ such that the following is true:
$$\forall x \in \mathbb{R}, |x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \delta_2\tag{1b}\label{1b}$$
Let $x \in \mathbb{R}$.
We will use the same $x$ in \eqref{1b} to obtain:
$$ |x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \delta_2\tag{1c}\label{1c}$$
Let $\delta = \delta_1$.
We assume $ |x – a| < \delta = \delta_1$.
We want to show:
$$ |g(f(x)) – g(f(a))| < \epsilon $$
By our assumption and \eqref{1c}, we know:
$$|f(x) – f(a)| < \delta_2 \tag{1d}\label{1d}$$
We substitute $f(x)$ for $x$ in \eqref{2b} to obtain:
$$|f(x) – f(a)| < \delta_2 \Longrightarrow |g(f(x)) – g(f(a))| < \epsilon \tag{2c}\label{2c}$$
Finally, we substitute \eqref{1d} in \eqref{2c} to obtain our desired conclusion:
$$|g(f(x)) – g(f(a))| < \epsilon$$
$\blacksquare$
And here is where I'm stuck. All I need to show is $0 < |f(x) – f(a)| $, and I will be able to complete the proof. But I don't know how to show that. Any ideas?
Edit: Proof has been updated and completed! If there's anything that can be improved, please let me know.
Best Answer
Your definition of continuity is wrong here - it is just $|x-a|<\delta$, and not $0<|x-a|<\delta$. For continuity, the case $x=a$ is valid too.