# Uniform Convergence of $(\sin x)^n$

Is the sequence of functions $(\sin(x))^n$ uniformly convergent in $[0,\pi]$? Can you give me a hint or solution, I have already already prove that it is UC in $[0,1]$ but I don't know how to proceed in $[1,\pi]$.

• How did you show it for $[0, 1]$? – Dair Aug 16 '17 at 0:11
• $|(sin(x))^n|\leq |x|^n$ for $x\in [0,1]$ – Ragnar1204 Aug 16 '17 at 0:13
• @Ragnar1204 $|x|^n$ is not uniformly convergent on $[0,1]$ or even $[0,1)$. – stewbasic Aug 16 '17 at 3:09

Hint: find the pointwise limit of this sequence of functions on $[0,\pi]$. If a sequence is uniformly convergent, then its limit is a continuous function. Is it continuous in this case?
It is not uniformly convergent: At $\pi/2$ we have $\sin(\pi/2)=1$ and thus $$\sin(\pi/2)^n\to 1$$ What happens everywhere else where $\sin(x)<1$?