I want to prove that if $f$ is continuous over some interval $[z_1,z_2]$, then any horizontally translated and horizontally "compressed/stretched" version of $f$ (i.e. a function of the form $f(bx+c)$, for any $b \neq 0,c \in \mathbb R$) will also be continuous over the adjusted interval $[\frac{z_1-c}{b}, \frac{z_2-c}{b}]$ (for $b \gt 0$) or $[\frac{z_2-c}{b}, \frac{z_1-c}{b}]$ (for $b \lt 0$).
In order to prove this, I will make use of 3 lemmas:
For any $a, c , L \in \mathbb R$, we have the following (for multiplication lemmas, $c \neq 0$):
$(\dagger_1)$: $\displaystyle\lim_{x \to a} f(x)=L \leftrightarrow \displaystyle\lim_{x \to a-c}f(x+c)=L$
$(\dagger_2)$: $\displaystyle\lim_{x \to a} f(x)=L \leftrightarrow \displaystyle\lim_{x \to \frac{a}{c}} f(cx)=L$
$(\dagger_3)$: $\displaystyle\lim_{x+c \to a} f(x)=L \leftrightarrow \displaystyle\lim_{x \to a-c}f(x)=L$
Our goal is prove that for any $a \in [\frac{z_1-c}{b}, \frac{z_2-c}{b}]$, $\quad \displaystyle\lim_{x \to a}g(x)=g(a)$, where $g(x)=f(bx+c)$. Note that $g(a)= f(ab+c)$.
We will assume that $b \gt 0$ without loss of generality.
The proof works as follows:
Let $a \in [\frac{z_1-c}{b},\frac{z_2-c}{b}]$. This means that $\frac{z_1-c}{b} \leq a \leq \frac{z_1-c}{b}$. Rearranging terms, we have: $z_1\leq ab+c \leq z_2$. This means that $(ab+c) \in [z_1,z_2]$.Therefore, by assumption, $\displaystyle\lim_{x \to ab+c} f(x)=f(ab+c)$.
From $(\dagger_3)$, we have: $\displaystyle\lim_{x \to ab+c} f(x)=\displaystyle\lim_{x-c \to ab}f(x)$
From $(\dagger_2)$, we have: $\displaystyle\lim_{x-c \to ab}f(x)=\displaystyle\lim_{x-c \to a}f(bx)$
From $(\dagger_3)$, we have: $\displaystyle\lim_{x-c \to a}f(bx)=\displaystyle\lim_{x \to a+c}f(bx)$
From $(\dagger_1)$, we have: $\displaystyle\lim_{x \to a+c}f(bx) = \displaystyle\lim_{x \to a}f(bx+c)$
So $g(a)=f(ab+c)=\displaystyle\lim_{x \to ab+c} f(x)=\displaystyle\lim_{x \to a}f(bx+c)=\displaystyle\lim_{x \to a}g(x) \quad \square$
Does this proof look alright?
