Why is this famous proof of the chain rule called "technically incorrect" in this pdf? [duplicate]

Solution 1:

If $g'(x) = 0$, the proof fails, but as mentioned in the comments, this can be handled as a separate case.

A more serious issue is the calculation $$(f \circ g)'(x) \cdot \left(\frac{1}{g'(x)}\right) = \lim_{h \to 0}\left(\frac{f(g(x+h)) - f(g(x))}{h}\right)\cdot\left(\frac{h}{g(x+h)-g(x)}\right)$$ If we write $$a(h) = \frac{f(g(x+h))-f(g(x))}{h}$$ and $$b(h) = \frac{h}{g(x+h)-g(x)}$$ then the above calculation is the assertion that $$\lim_{h \to 0}\ a(h) \cdot \lim_{h \to 0}\ b(h) = \lim_{h \to 0}\ a(h)b(h),$$ in other words, that the limit of a product is the product of the limits. This is true, provided that both limits on the left-hand side exist. But the existence of $\displaystyle \lim_{h \to 0}\ a(h)$ is exactly what we are trying to prove, so the argument is circular.

Another issue worth noting is the final assertion, namely $$f'(g(x)) = \lim_{h \to 0}\frac{f(g(x+h)) - f(g(x))}{g(x+h)-g(x)}$$ In fact, the definition of the derivative of $f$, evaluated at $g(x)$, gives us $$f'(g(x)) = \lim_{k \to 0}\frac{f(g(x)+k) - f(g(x))}{k}$$ The only reason we are able to conclude that these two expressions are equal (even after handling the $g(x+h) = g(x)$ case properly) is because $g$ is continuous at $x$. This of course follows from the differentiability of $g$ at $x$, but a careful proof would point this out.

Solution 2:

What if $g$ is a constant function?