Chain rule proof assumption.

Question

I am reading this in my text:

I have 3 questions:

1. What is $\epsilon$.

2. "If we define $\epsilon$ to be 0 when x is 0, then $\epsilon$ becomes a continuous function of x." Why is this true? I don't see it.

3. Why is $\Delta{y} = f'(a) \Delta{x} + \epsilon \Delta{x} $ ? Isn't the change in y just = $f'(a) \epsilon{x}$?

Answer 1

1. The answer is in the text: $\varepsilon$ is by definition the difference $$\dfrac{\Delta y}{\Delta x}-f^\prime(a)$$ Note that this makes $\varepsilon$ a function of $\Delta x$, in particular (in the same way as $f(x)=\dfrac{2}{x}-3$ is a continuous function of $x$ away from $x=0$).
2. Note that the above assumption makes no sense when $\Delta x=0$. Therefore you need a separate definition in that case. But one choice is better than all others because it will make $\varepsilon$ a continuous function of $\Delta x$.
3. This is simply rewriting the definition of $\varepsilon$.

Answer 2

1) $\epsilon =\frac {\triangle y}{\triangle x} - f'(a)$.

This is a value dependent upon $\triangle x$ and $\triangle y$. $\triangle y$ is dependent up $\triangle x$. So we can think of $\epsilon$ as a function of $\triangle x$.

IMO, I'd find it clearer if we wrote $\epsilon$ as $\epsilon_{\triangle x}$ or as $\epsilon(\triangle x)$. But this text didn't. Oh, well.

Notice: $\epsilon_{\triangle x}$ is undefined at $\triangle x = 0$.

And notice $\lim_{\triangle x\to 0} \epsilon_{\triangle x}=\lim_{\triangle x\to 0} (\frac {\triangle y}{\triangle x} - f'(a))=0$.

2) A function $g(x)$ is continuous and $x = a$ if $\lim_{x\to a} g(x) = g(a)$. That's practically the definition of continuous.

So we define $\epsilon_{\triangle x}$ as: if $\triangle x \ne 0$ then $\epsilon_{\triangle x}=\frac {\triangle y}{\triangle x} - f'(a)$; if $\triangle x = 0$ then $\epsilon_{\triangle x}=0$.

Now $\lim_{\triangle x\to 0} \epsilon_{\triangle x}=\lim_{\triangle x\to 0} (\frac {\triangle y}{\triangle x} - f'(a))=0 = \epsilon_0$.

So $\epsilon_{\triangle x}$ is a continuous function at $\triangle x = 0$

3) "Why is Δy=f′(a)Δx+ϵΔx ? Isn't the change in y just = f′(a)ϵx?"

No, because $\epsilon$ is not a constant.

$\epsilon = \frac {\Delta y}{\Delta x} - f'(a)$ so

$\epsilon \Delta x = \Delta y - f'(a)\Delta x$

$\Delta y = f'(a)\Delta x + \epsilon \Delta x$.

This, however, is differentiable because $\epsilon$ is a continuous function.