I have heard it been said that the single-variable chain rule can be proven using the notion that the derivative gives the "best" linear approximation of a function around a point, in the sense that if $f$ is differentiable at $x$, then $f'(x)$ is the unique number such that
$$
\lim_{\Delta x \to 0}\frac{\bigl(f(x+\Delta x)-f(x)\bigr)-f'(x)\Delta x}{\Delta x}=0 \, .
$$
This means that we can write
$$
f(x+\Delta x) = f(x) + f'(x)\Delta x + r(\Delta x) \, ,
$$
where the remainder term $r(\Delta x)$ is "small" in the sense that $\lim_{\Delta x \to 0}\frac{r(\Delta x)}{\Delta x}=0$. I have never read a calculus book that uses this idea to prove the chain rule, and so I tried to come up with my own proof. Is it correct?
THEOREM: The chain rule. Suppose that $g$ is differentiable at $x$, and $f$ is differentiable at $g(x)$. Then, $f \circ g$ is differentiable at $x$, and
$$
(f \circ g)'(x)=f'(g(x)) \cdot g'(x) \, .
$$
The basic idea behind my purported proof is that
$$
f(g(x+\Delta x))\approx f(g(x)+g'(x)\Delta x) \approx f(g(x))+f'(g(x))g'(x)\Delta x \, .
$$
First, we will make the following definition out of convenience:
DEFINITION: We say that a remainder $r(t)$ is small as $t\to0$ if $\lim_{t \to 0}\frac{r(t)}{t}=0$.
Since $g$ is differentiable at $x$,
$$
\lim_{\Delta x \to 0}\frac{\bigl(g(x+\Delta x)-g(x)\bigr)-g'(x)\Delta x}{\Delta x} = 0 \, .
$$
If we define $r_1(\Delta x)=\bigl(g(x+\Delta x)-g(x)\bigr)-g'(x)\Delta x$, then $r_1(\Delta x)$ is small as $\Delta x \to 0$.
In a similar fashion, $f$ is differentiable at $g(x)$, and so
$$
\lim_{h \to 0}\frac{\bigl(f(g(x)+h)-f(g(x))\bigr)-f'(g(x))h}{h}=0 \, .
$$
If we define $r_2(h)=\bigl(f(g(x)+h)-f(g(x))\bigr)-f'(g(x))h$, then $r_2(h)$ is small as $h\to0$. By keeping track of the remainders $r_1$ and $r_2$, we can show that $\bigl((f \circ g)(x+\Delta x)-(f \circ g)(x)\bigr)-f'(g(x))g'(x)\Delta x$ is small as $\Delta x \to 0$, and thus $(f \circ g)'(x) = f'(g(x)) \cdot g'(x)$. For ease of notation, let $\Delta u = g'(x)\Delta x+r_1(\Delta x)$. Then,
\begin{align}
f(g(x+\Delta x)) &= f(g(x)+\Delta u) \\[4pt]
&= f(g(x)) + f'(g(x))\Delta u + r_2(\Delta u) \\[4pt]
&= f(g(x)) + f'(g(x))g'(x)\Delta x + f'(g(x))r_1(\Delta x)+r_2(\Delta u) \, .
\end{align}
We are left with two remainder terms: $f'(g(x))r_1(\Delta x)$ and $r_2(\Delta u)$. The first is obviously small as $\Delta x \to 0$. To show that $r_2(\Delta u)$ is small as $\Delta x \to 0$, define the functions $\psi$ and $\phi$:
\begin{align}
\psi(\Delta x) &= \Delta u = g'(x)\Delta x+r_1(\Delta x) \, ,\\[5pt]
\phi(h)&=
\begin{cases}
\dfrac{r_2(h)}{h} &\text{if $h\neq0$} \\
0 &\text{if $h=0$.}
\end{cases}
\end{align}
It is easy to show that $\psi$ is continuous at $0$, and that $\phi$ is continuous at $\psi(0)=0$. Hence, $\phi \circ \psi$ is continuous at $0$, meaning that
$$
\lim_{\Delta x \to 0}\phi(\psi(\Delta x))=\lim_{\Delta x \to 0}\frac{r_2(\psi(\Delta x))}{\psi(\Delta x)}=\lim_{\Delta x \to 0}\frac{r_2(\Delta u)}{\Delta u} = 0 \, .
$$
This completes the proof.
Best Answer
Your proof works, but it could be made much cleaner (and closer to your "basic idea") with the help of big O notation. In particular, we will say that $r(\Delta x) = o(\Delta x)$ if $$ \lim_{\Delta x \to 0} \frac{r(\Delta x)}{\Delta x} = 0. $$ Moreover, to make everything easier to type, I will write $h$ instead of $\Delta x$.
Note that $f'(x)$ is the unique function for which $f(x + h) = f(x) + f'(x) h + o(h)$ (and may be defined as such). With that, we note that \begin{align} f(g(x + h)) &= f(g(x) + g'(x)h + o(h)) \\ &= f(g(x)) + f'(g(x))(g'(x)h + o(h))h + o(h) \\ &= f(g(x)) + f'(g(x))g'(x) h + [f'(g(x))o(h) h + o(h)] \\ & = f(g(x)) + f'(g(x))g'(x) \cdot h + o(h). \end{align} Note that the statement $$ f'(g(x))o(h) h + o(h) = o(h) $$ is shorthand for your statement regarding limits. By our "definition" of the derivative, we conclude that $f'(g(x))g'(x)$ is the derivative of $f \circ g$ at $x$, which is what we wanted.