The continuity law for composition states, informally, that:

$ $ \text{IF } f \text{ is continuous at } g \ \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(x)} 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}, 0<|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}, 0<|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}, 0<|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}, 0<|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}, 0<|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}, 0<|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}, 0<|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:

$ $ 0 < |x – a| < \delta_1 \Longrightarrow |f(x) – f(a)| < \delta_2\tag{1c}\label{1c}$ $

Let $ \delta = \delta_1$ .

We assume $ 0 < |x – a| < \delta = \delta_1$ .

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:

$ $ 0<|f(x) – f(a)| < \delta_2 \Longrightarrow |g(f(x)) – g(f(a))| < \epsilon \tag{2c}\label{2c}$ $

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?