Given the set of all real number $\mathbb R$ and a real function $g:\mathbb R \rightarrow \mathbb R =\begin{cases}x^n&\textrm{if } x\geq0 \\0 &\textrm{if } x=0)\end{cases}$
Determine all the natural number $n$ so that $\frac{dg(0)}{dx}$ does not exist.
I read in a literature that "If $f$ is not continuous at $a$, then $f'(a)$ does not exist."
Since $g=\begin{cases}x^n& \textrm{if }x\geq0 \\0 &{\textrm{if } x=0}\end{cases}$, the first derivative is $\frac{dg(x)}{dx}=\begin{cases}nx^{n-1}&\textrm{if } x\geq0 \\0 &{\textrm{if }x=0}\end{cases}$
I try to find all natural number n so that $g(x)$ is not continuous at $x=0$ so that $\frac{dg(0)}{dx}$ does not exist.
I also read that $f$ is continuous at $c$ if and only if
1. $\lim_{x\rightarrow c}f(x)$ exists,
2. $f(c)$ exists, and
3. $\lim_{x\rightarrow c}f(x)=f(c)$.
I have checked for $g(x)$
1. $\lim_{x\rightarrow 0}g(x)=\lim_{x\rightarrow 0}x^n=0 \forall n \in \mathbb N$
2. $g(0)=0$
3. $\lim_{x\rightarrow 0}g(x)=g(0)=0$.
Thus, $g(x)$ is continuous at $0 \forall n \in \mathbb N$
This makes me conclude that there is no natural number $n$ so that $\frac{dg(0)}{dx}$ does not exist. Am I wrong?
$\endgroup$ 51 Answer
$\begingroup$If $n=1$ then $$\lim_{\epsilon\downarrow0}\frac{g\left(\epsilon\right)-g\left(0\right)}{\epsilon}=1\neq0=\lim_{\epsilon\uparrow0}\frac{g\left(\epsilon\right)-g\left(0\right)}{\epsilon}$$ This means that $$\lim_{\epsilon\rightarrow0}\frac{g\left(\epsilon\right)-g\left(0\right)}{\epsilon}$$ is not defined.
If $n>1$ then $$\lim_{\epsilon\downarrow0}\frac{g\left(\epsilon\right)-g\left(0\right)}{\epsilon}=0=\lim_{\epsilon\uparrow0}\frac{g\left(\epsilon\right)-g\left(0\right)}{\epsilon}$$ Wich means that $$\lim_{\epsilon\rightarrow0}\frac{g\left(\epsilon\right)-g\left(0\right)}{\epsilon}$$ is defined and equals $0$.
If a function is not continuous at some $a$ then it is not differentiable at $a$. In general the converse of this is not true (see comment of Henning for a counterexample).
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