Help with proof of chain rule

I'm following along on this proof of the chain rule. All is clear to me except the step where they say:

Differentiablility implies continuity; therefore $\Delta_u \to 0$ as $\Delta_x \to 0$ in $f\big(g(x)\big)$.

Could someone explain to a calculus newb why that's the case, in simple terms?

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2 Answers

A piece of advice: try another site as the one you linked uses what I consider pretty poor notation.

If we have $\,f(g(x))\,$ then I'd go

$$\lim_{x\to x_0}\frac{f(g(x))-f(g(x_0))}{x-x_0}=\lim_{x\to x_o}\frac{f(g(x))-f(g(x_0))}{g(x)-g(x_0)}\frac{g(x)-g(x_0)}{x-x_0}$$

And what that site says with its cumbersome (for me) notation is that $\,x\to x_0\Longrightarrow g(x)\to g(x_0)\,$ , since differentiability implies continuity.

Some care must be put to avoid the possibility of dividing by zero in the above, of course.

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(before answering your question, let me tell you that I don't really recommend that proof; the main problem is that derivatives are calculated at a point, and the notation used in the proof obscures that fact completely)

You are calculating your derivatives at some $x=x_0$. Then $\Delta x=x-x_0$; and $\Delta u=u(x)-u(x_0)$. As $u$ is differentiable, it is continuous. Continuity at $x_0$ for $u$ means exactly that $u(x)-u(x_0)\to0$ when $x-x_0\to0$, i.e. $\Delta u\to 0$ when $\Delta x\to 0$.

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