Evaluate $\lim_{x \to 0}{\frac{\sqrt[3]{1 + cx} - 1}{x}}$

Question

This problem is a worked example in James Stewart's Calculus: Early Transcendentals [8E], page 169.

Find: $$\lim_{x \to 0}{\frac{\sqrt[3]{1 + cx} - 1}{x}}$$

where $c$ is a constant. The answer given is $\frac{c}{3}$, and I understand the derivation, but it seems that the answer should be that the limit does not exist, because of this earlier statement in the book about the composition of limits:

If $f$ is continuous at $b$, and $\lim_{x \to a}{g(x)} = b$, then $\lim_{x \to a}{f(g(x))} = f(b) = f(\lim_{x \to a}{g(x)})$

Let $f(x) = \frac{c(x - 1)}{x^3 - 1}$ and $g(x) = \sqrt[3]{1 + cx}$. Then, $f(g(x)) = \frac{\sqrt[3]{1 + cx} - 1}{x}$. Therefore, if $\lim_{x \to 0}{f(g(x))}$ exists, then $f(1)$ should be continuous according to this statement, because $\lim_{x \to 0}{g(x)} = 1$. But to be continuous, $f(1)$ should be defined, which it is not since $f(1)$ makes a zero denominator.

I'm sure that I have some misunderstanding but I'm not sure what it is.

Answer 1

If $f$ is continuous at $b$ … then …

You're trying to applying the theorem backwards like concluding that a duck is a chicken from “If $X$ is a chicken, then $X$ has two legs” and “for every $X$, if $X$ is a duck, then $X$ has two legs”.

Answer 2

If $f$ is not defined at $b = 1$ then it is certainely not continuous at $b = 1$, so the conditions of your statement don't hold.

Note that $f$ can be defined at 1 by

$$\lim_{x \to 1} \frac{c(x-1)}{(x-1)^3} = \lim_{x \to 1} \frac c{x^2+x+1} = \frac c3$$

And this makes $f$ continuous at $1$

Answer 3

Your $f$ is not continuous at $1$ because you made it so and as a consequence you can not apply the statement at all (you don't verify the hypotheses).

But nothing forbids to define $f(1):=\frac c3$, which makes the function continuous (because $\lim_{g\to1}f(g)=\frac c3$) and allows to use the statement.