Let f(x) be x^1/3. Show that f '(0) does not exist.

Question

I am new in calculus and we will have our first test on Monday from chapter 1 and 2. I have two questions that I don't know the right answer but tried to do them as properly as I could. I want to do good on the test so can someone tell me how to do better in these kind of proof questions?

Let f(x) be $\sqrt[3]{x}$.

(a) Show that $f '(0)$ does not exist.

(b) show that $\sqrt[3]{x}$ has a vertical line at $(0,0)$

For (a) I know that when we take the derivative of $\sqrt[3]{x}$ we get $\frac{1}{3\sqrt[3]{x^2}}$ and when we plug $0$ for $x$ we get $\frac{1}{0}$ which is not defined thus, it does not exist. However I think the question wants me to solve it in a more logical way that includes the use of limits.

For (b) I say; by the definition of the derivative the equation of the tangent line is $y=0$ because the slope is $0$ at that point. Therefore, it creates a vertical line.

and in a mathematical way can I just say; $$\lim _{x\to 0^+}\left(\frac{1}{3\sqrt[3]{x^2}}\right)=\infty$$ $$\lim _{x\to 0^-}\left(\frac{1}{3\sqrt[3]{x^2}}\right)=\infty$$

both left and right limits are equal and goes to infinity, so there must exist a vertical line?

Answer 1

For part (a), you must invoke the definition of a differential, the first principle of differentiation, which states that $f(x)$ is differentiable at $x=a$ iff $$\lim_{h \to 0^-}\frac{f(a+h)-f(a)}{h} = \lim_{h \to 0^+}\frac{f(a+h)-f(a)}{h} = f'(a)$$

So, it suffices to show that the LHS and RHS are not equal above, to show that $f'(x)$ does not exist at $x = a$, here $f(x) = \sqrt[3] x$, and $a = 0$.

Answer 2

1. The domain of $f$ is $[0, \infty)$.

2. $\frac{f(x)-f(0)}{x-0}=\frac{1}{x^{2/3}} \to \infty$ for $x \to 0+0$