How do I check if x^3 is differential at 0?

Question

If I were to change the function of f(x) to $$f(x) = x^3$$

Can I rewrite the limit to this, by substituting the function of $$x^3$$ directly?

$$\lim_{x\to 0} {x^3-0\over x-0}$$

as compared to

Answer 1

We have

$\lim\limits_{x\to 0^+} f{'}(x)= \lim\limits_{x\to 0^+}1/3\cdot x^{-2/3}=+\infty \ $

$\lim\limits_{x\to 0^-} f{'}(x)= \lim\limits_{x\to 0^-}1/3\cdot x^{-2/3}=-\infty$

Thus $\lim\limits_{x\to 0^+} f{'}(x)\color{red}{\neq} \lim\limits_{x\to 0^-} f{'}(x)$

That means that $f'(x)$ is undefined at $x=0$.

See the graph at wolfram alpha

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In contrast to $g(x)=x^3$

$\lim\limits_{x\to 0^+} g{'}(x)= \lim\limits_{x\to 0^+}3x^2=0$

$\lim\limits_{x\to 0^-} g{'}(x)= \lim\limits_{x\to 0^-}3x^2=0$

$\lim\limits_{x\to 0^+} g{'}(x)\color{blue}{=} \lim\limits_{x\to 0^-} g{'}(x)$

Answer 2

Yes, of course. We have $$f'(x) = 3x^2$$ which is defined on $\mathbb{R}$ as is any polynomial function.

Answer 3

You may write

$$\lim_{x\to c}\frac{x^{1/3}-c^{1/3}}{x-c}=\lim_{t^3\to c^3}\frac{t-c}{t^3-c^3}=\frac1{3t^2}=\frac1{3x^{2/3}}$$

which is of course undefined at $x=0$.