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Let $k∈[0,∞)$ be a real number. Define
$f_k(t)$=$t^k$sin($1/t$) ,if $t≠0$ and $0$ if $t=0$ . Let $A$={$k∈[0,∞) : f_k$ is differentiable} . then $A$=?

i am not sure how to solve this problem.can anybody help me.

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    There is no doubt that $t^k\sin(1/t)$ is *right-differentiable* at $t=0$ for all $k\gt 1$. And under the usual conventions, it is differentiable at $t=0$ for integers $x\gt 1$. One can extend this to $k=p/q\gt 1$, where $p$ and $q$ are relatively prime positive integers, and $q$ is not divisible by $2$. But for general $k\gt 1$, like $k=\sqrt{2}$, there is the issue that $t^k$ is not defined for negative $t$.2012-12-31

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Hint: The function $f_k(t)=t^k\sin(1/t)$ will be differentiable for all $k\geq 0$ when $t>0$, so the only potential problem is $t=0$. At that point, use the definition of the derivative. $f_k^{'}(0)=\lim_{h\rightarrow0}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0}\frac{h^{k}\sin\left(\frac{1}{h}\right)}{h}.$ So for which $k$ does $\lim_{h\rightarrow0}h^{k-1}\sin\left(\frac{1}{h}\right)$ exist?

Note: I am assuming that $f_k:[0,\infty)\rightarrow \mathbb{R}$, and that we do not define $f_k$ on the negative integers. This is because, as André Nicolas points out, for negative $t$, there is not clear definition of that $t^k$ means for every $k$.