Recently I was going through the chain-rule derivative proof in Stewart Calculus-
It's a great approach to avoid the objections of $\Delta$$u$$=0$. But I don't understand why do we need to prove that $\epsilon$ is a continuous function of$\Delta$$x$. The book claims it to be a property of a differentiable function, but what is the significance of this property and how are we helped here by using it? Also, does it really help in some later stages of calculus? If so, then how? It'd be really helpful if someone pointed out why it was necessary to use $\epsilon$ as a continuous function here.
Notice that here we are using that $y=f(x)$ and increment of $y$ is defined by $$ \Delta y=f(a+\Delta x)-f(a). $$ By the definition of derivative $$ f'(a)=\lim\limits_{\Delta x \to 0}\frac{\Delta y}{\Delta x}. $$ Defining $\epsilon$ as $$ \epsilon=\frac{\Delta y}{\Delta x}-f'(a) $$ for $\Delta x \neq 0$ we have that $$ \lim\limits_{\Delta x \to 0} \epsilon=\lim\limits_{\Delta x \to 0}\frac{\Delta y}{\Delta x}-f'(a)=0. $$ Define then $\epsilon$ to be $0$ when $\Delta x=0$. If we consider $\epsilon=\epsilon(\Delta x)$ as a function of $\Delta x$ we have that $\epsilon$ is continuous. In fact, notice that the only point that continuity might fail is at $\Delta x=0$ but as we defined $\epsilon(\Delta x)$ $$ \lim\limits_{\Delta x \to 0} \epsilon=0=\epsilon(0). $$ Hence $\epsilon$ is in fact continuous with respect to $\Delta x$. This property is what allows you to write equations 8 and 9 where $\epsilon_1 \to 0$ as $\Delta x \to 0$ and $\epsilon_2 \to 0$ as $\Delta u \to 0$.
Notice that what this property states is that if $f$ is differentiable at $a$ then you can find a continuous function on $\Delta x$ here denoted by $\epsilon$ such that $$ f(a+\Delta x)-f(a)=f'(a) \Delta x +\epsilon \Delta x $$ where $\epsilon \to 0$ as $\Delta x \to 0$. Notice that the converse is also true. If there is a function $\epsilon$ of $\Delta x$ and a constant $b$ such that $$ f(a+\Delta x)-f(a)=b \Delta x +\epsilon \Delta x $$ where $\epsilon \to 0$ as $\Delta x \to 0$ then $f$ is differentiable at $a$ and $f'(a)=b$. In fact, we have that for $\Delta x \neq 0$ $$ \frac{f(a+\Delta x)-f(a)}{\Delta x}=b+\epsilon $$ and $$ \lim\limits_{\Delta x \to 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}=\lim\limits_{\Delta x \to 0}b+\epsilon=b+\lim\limits_{\Delta x \to 0}\epsilon=b+0=b. $$ Hence by definition $f$ is differentiable at $a$ and $f'(a)=b$.