Definition of limit with$ f(x)=|x^3|$

Question

Using the definition of the limit I tried to find the derivative of $f(x)=|x^3|$. I came up with: $$f'(x)=\frac{3x^5}{|x^3|}$$

Question: Why is the derivative (according to this answer) not defined for $x=0$ (division by zero) while $f'(0)$ actually exists, since: $$f'(0)=\lim_{x\to 0}\frac{|x^3|-|0^3|}{x-0}=0$$

Elaboration using the definition of the limit:

$$f'(x)=\lim_{h\to 0}\frac{|(x+h)^3|-|x^3|}{h}$$ $$=\lim_{h\to 0}\frac{\sqrt{(x+h)^6}-\sqrt{x^6}}{h}$$ $$=\lim_{h\to 0}\frac{(x+h)^6-x^6}{h(\sqrt{(x+h)^6}+\sqrt{x^6})}$$ $$=\lim_{h\to 0}\frac{h^5 + 6 h^4 x + 15 h^3 x^2 + 20 h^2 x^3 + 15 h x^4 + 6 x^5}{(\sqrt{(x+h)^6}+\sqrt{x^6})}$$ $$=\frac{6x^5}{2\sqrt{x^6}}=\frac{3x^5}{|x^3|}$$ Which is not defined for $x=0$.

Answer 1

Consider the limit from below ($x$ tending from negative to zero) and from above separately. They both evaluate to zero.

Answer 2

Your problem was saying that

$$=\lim_{h\to 0}\frac{h^5 + 6 h^4 x + 15 h^3 x^2 + 20 h^2 x^3 + 15 h x^4 + 6 x^5}{(\sqrt{(x+h)^6}+\sqrt{x^6})}=\lim_{h\to 0}\frac{6x^5}{2\sqrt{x^6}}$$

which is not true. If you set $h$ to $0$, then you must set all values of $h$ to $0$, and drop the limit all together. You cannot just set some of them to $0$. For exapmle, you cannot say

$$\lim_{h\to 0} \frac hh = \lim_{h\to 0} \frac0h = 0$$ because the first equality is not true.

What you get in your limit calculation is

$$=\lim_{h\to 0}\frac{h^5 + 6 h^4 x + 15 h^3 x^2 + 20 h^2 x^3 + 15 h x^4 + 6 x^5}{(\sqrt{(x+h)^6}+\sqrt{x^6})}=\\=\lim_{h\to 0}\frac{6x^5}{\sqrt{(x+h)^6}+\sqrt{x^6}}+\frac{h(h^4 + 6 h^3 x + 15 h^2 x^2 + 20 h^1 x^3 + 15 x^4)}{\sqrt{(x+h)^6}+\sqrt{x^6}}$$

Which is equal to the sum of limits of both sumands, if both limits exist.

Now, you can easily see that

$$\lim_{h\to0} \frac{6x^5}{\sqrt{(x+h)^6}+\sqrt{x^6}} = 3\mathrm{sign}(x)x^2$$ no matter what the value of $x$ is. If $x=0$, the limit becomes $$\lim_{h\to0} \frac{0}{\sqrt{h^6}}$$ and is equal to $0=3\cdot 0^2.$

For the second limit, you can also quickly see that it is $0$ no matter what the value of $x$ is. If $x\neq 0$, then the result is obvious, if $x=0$, then the limit is $$\lim_{h\to0}\frac{h^5}{\sqrt{h^6}}=\lim_{h\to 0} h^2\cdot\mathrm{sign}(h)=0$$

Answer 3

Let's tackle first the problem for $x=0$. Observe that $|a^3|=a^2\,|a|$, so $$ f'(0)=\lim_{h\to0}\frac{|(0+h)^3|-0}{h}= \lim_{h\to0}\frac{|h^3|}{h}= \lim_{h\to0}\frac{h^2\,|h|}{h}=\lim_{h\to0}h\,|h|=0 $$

For $x\ne0$: \begin{align} f'(x) &=\lim_{h\to0}\frac{\sqrt{(x+h)^6}-\sqrt{x^6}}{h}\\[6px] &=\lim_{h\to0}\frac{(x+h)^6-x^6}{h(\sqrt{(x+h)^6}+\sqrt{x^6})}\\[6px] &=\lim_{h\to0}\frac{h(6x^5+hp(x,h))}{h(\sqrt{(x+h)^6}+\sqrt{x^6})}\\[6px] &=\lim_{h\to0}\frac{6x^5+hp(x,h)}{\sqrt{(x+h)^6}+\sqrt{x^6}}\\[6px] &=\frac{6x^5}{2|x|^3}=\frac{3x^5}{x^2|x|}=\frac{3x^3}{|x|} \end{align} (where $p(x,h)$ is some polynomial in $x$ and $h$ that you can compute explicitly).