I am preparing for my exam and need help with the following task:
Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by $f(x)= \begin{cases}1, x\geq0 \\ 0 , x<0 \end{cases}$. Let $g_k$:$\mathbb{R}\to\mathbb{R}$ be a function with $g_k(x)=(sin(x))^k\cdot f(x) $ $\forall k\in \mathbb{N}$(without $0$). For which $k\in\mathbb{N}$
- a) $g_k$ is continuous in $x=0$ ?
- b) $g_k$ is differentiable in $x=0$ ?
a) $g_k$ is continuous, if $\lim\limits_{x \rightarrow 0}{g_k(x)}$$=g_k(0)=0$. So we have to check $\lim\limits_{x \uparrow 0}{g_k(x)}$ and $\lim\limits_{x \downarrow 0}{g_k(x)}$.
$\lim\limits_{x \uparrow 0}$ $(sin(x))^k$ $\cdot f(x)$ =$0$ Thats because $f(x)$ stays $0$.
$\lim\limits_{x \downarrow 0}$ $(sin(x))^k$ $\cdot f(x)$=$0$ too. Although $f(x)$ stays 1, $\lim\limits_{x \rightarrow 0}$ $(sin(x))^k$=$0$. Thats because $sin(x) \to 0$ and we can compare it with $(\frac{1}{x})^k$ with k>0, that is converging to $0$ as well. Thats why $g_k(x)$ is continuos in $x=0$ $\forall k\in\mathbb{N}$. Is this proof correct?
b) $g_k$ is differentiable in $x=0$, if $\lim\limits_{h \rightarrow 0}$ $\frac{(sin(h))^k\cdot f(h)}{h}$ and thats where I am stuck now. I thought about l'hopital's rule, but we only can use it, if $(sin(h)^k\cdot f(h)$ is differentiable. Thats the condition for l'hopital. Obviously the result then has to be $\lim\limits_{h \rightarrow 0}$ ${k(sin(h))^{k-1}\cdot f(h)}$, which will lead us to the result, that k>1. If k<1, then we would have $$\lim\limits_{h \downarrow 0} {k(sin(h))^{k-1}\cdot f(h)}=\infty$$ If we have k=1, then we have $$\lim\limits_{h \downarrow 0} { f(h)}=1\neq 0= \lim\limits_{h \uparrow 0} { f(h)}$$
So how am I gonna solve this $\lim\limits_{h \rightarrow 0}$ $\frac{(sin(h))^k\cdot f(h)}{h}$ pretty easily.
Is there anyone who could give me an advice? I would be very grateful.