This post will be broken up into two sections: the first section will contain the proof, and the second section will contain the question. The proof will be written formally, as the question is more easily understood referencing the formal description. (Note, the context of this post is in Spivak's Calculus, which treats all functions, unless otherwise stated, as having a domain of $\mathbb R$).
Prove: $\displaystyle \lim_{h \to 0}f(ch)=\displaystyle \lim_{ch \to 0}f(ch)$ for $c\neq 0$, which is equivalent to:
If $c\neq 0$ and $\displaystyle \lim_{h \to 0}f(ch)=L$, then $\displaystyle \lim_{ch \to 0}f(ch)=L$ $\quad$ and $\quad$ If $c\neq 0$ and $\displaystyle \lim_{ch \to 0}f(ch)=L$, then $\displaystyle \lim_{h \to 0}f(ch)=L$
We will only prove the first implication (the converse is completed similarly):
By assumption: $\displaystyle \lim_{h \to 0}f(ch)=L \iff \forall \varepsilon \gt 0 \ \exists \delta \gt 0 \ \forall h \in \mathbb R \left [ 0 \lt |h| \lt \delta \rightarrow |f(ch)-L| \lt \varepsilon \right ]$
We want to show that for an arbitrary $\varepsilon$, we can construct a $\delta$ such that $\color{red}{\forall ch} \in \mathbb R \left [ 0 \lt |ch| \lt \delta \rightarrow |f(ch)-L| \lt \varepsilon \right]$
For $\varepsilon$, we know by assumption that there is a $\delta_{\varepsilon}$ such that:
$\forall h \in \mathbb R \left [ 0 \lt |h| \lt \delta_{\varepsilon} \rightarrow |f(ch)-L| \lt \varepsilon \right ]$
Now, consider a $\delta ^* = \min\left(\delta_{\varepsilon},\frac{\delta_{\varepsilon}}{|c|}\right)$.
If $0 \lt |h| \lt \delta^* \leq \delta_{\varepsilon}$, by assumption we have: $|f(ch)-L| \lt \varepsilon$.
Further, if $0 \lt |h| \lt \delta^* \leq \frac{\delta_{\varepsilon}}{|c|}$, then $0 \lt |ch| \lt |c|\delta^*$.
Therefore, let our desired $\delta$ be defined as $\delta = |c|\delta^*$. As long as $0\lt|ch| \lt \delta$, all of our criteria is met.
In the above proof, I made use of the following statement:
$\color{red}{\forall ch} \in \mathbb R \left [ 0 \lt |ch| \lt \delta \rightarrow |f(ch)-L| \lt \varepsilon \right]$
Through my brief experience in maths, the universally quantified object $ch$ is atypical. I suspect the proper way to denote this is by establishing a function of the form: $g(h)=ch$ and then defining a single symbol as representing its output. i.e. something like $s_h :=g(h)$. More specifically, we should write $g$ formally as: $g: \mathbb R \to \mathbb R$ where $h \mapsto ch$.
We would then rewrite the statement as:
$\color{red}{\forall s_h} \in \mathbb R \left [ 0 \lt |s_h| \lt \delta \rightarrow |f(s_h)-L| \lt \varepsilon \right]$
This seems to emulate the more familiar notation of a universal quantifier, where only one symbol follows the quantifier.
I do not really know the deep theory behind first-order logic, but I suspect the reason this proof "works out" is because my new symbol $s_h$ has the capacity of sweeping through all objects within $\mathbb R$. Said differently, the previously defined function $g$ can be shown to be surjective with respect to $\mathbb R$.
If the above is true, are there times where the change of variable function is not surjective, and this causes the equality between two limits to fail?
Best Answer
In your example, $\varphi(t) = ct$ works since if $$\lim\limits_{h \to h_0} f(ch) = \ell$$ exists, then we can take $t_0 = h_0/c$ (legal because $c \ne 0)$ so that $$\lim\limits_{t \to t_0} \varphi(t) = \lim\limits_{t \to h_0/c}(ct) = c \cdot \dfrac{h_0}{c} = h_0$$
For counter-example, we should think of any example that at least one of those conditions is not satisfied. Think of $$\lim\limits_{x\to 0} x$$ (where $f(x) = x$ and $x_0 = 0$) for simplicity. Then, if we take $\varphi(t) = t^2+1$, for example, we cannot find $t_0 \in \mathbb R \cup \{\pm\infty\}$ such that $$\lim\limits_{t\to t_0} \varphi(t) = 0$$ (which is our $x_0$) since $\varphi(t) \ge 1$ for any $t \in \mathbb R$ and also $$\lim\limits_{t \to \pm\infty} \varphi(t) = +\infty \ne 0$$ Note that, however, $\varphi(t) = 1/t$ would work since we can take $t_0 = \infty$ so that $$\lim\limits_{t \to t_0} \varphi(t) = 0 = x_0$$ and $$\begin{array}{c|c} \varphi : & \begin{matrix}\mathbb R \longrightarrow \mathbb R \\ t \longmapsto \dfrac{1}{t}\end{matrix} \end{array}$$ is not surjective.
EDIT: I thought you were mainly asking for change of variables inside limits and the importance of surjectivity in the substitution.
According to reasonable comments, let's prove the following with the abovementioned arguments and the definition of limits: $$\color{navy}{\lim_{h \to 0}f(ch)}= \ell \iff \color{fuchsia}{\lim_{ch \to 0}f(ch)} = \ell \tag{$c \ne 0$}$$ $(\Leftarrow)$ In the second limit, take $\varphi(t) = t/c$. Then ($t \leftrightarrow h$) $$\color{fuchsia}{\lim_{h \to 0} f(h)} = \ell$$ which means for any $\varepsilon > 0$ we have $$|f(h) - \ell| < \varepsilon \tag{$\star$}$$ whenever $|h| < \color{fuchsia}\delta$ for some $\color{fuchsia}\delta = \color{fuchsia}\delta(\varepsilon) > 0$. Then we can take $\color{navy}{\delta'} = \color{fuchsia}\delta/|c|$ so that $(\star)$ would imply $$|f(ch) - \ell| < \varepsilon$$ whenever $|h| < \color{navy}{\delta'} = \color{fuchsia}\delta/|c|$ since $|ch| < |c|\color{navy}{\delta'} = \color{fuchsia}\delta$. So, we have $$\lim_{h \to 0}f(ch) = \ell$$ The converse is similar.