Showing $\sin^n(x)$ does not converge uniformly on $[0,\pi/2]$ I just want to check that I am correct in my argument that $f_n(x) = \sin^n(x)$ does not converge uniformly.
When $x = \pi/2$, $\sin^n(x) = 1$ for all $n$, hence $ f_n(\pi/2) \rightarrow 1$, However, for all other $x$, $f_n(x) \rightarrow 0$. Hence the limit function of a sequence of continuous function is not continuous, and so the convergence is not uniform.
 A: Correct. You can also argue as follows: Let $f(x)=0$ for $x\in [0,\pi /2)$ and $f(\pi /2)=1.$ Each $f_n$ is continuous with $f_n(0)=0$ and $f_n(\pi /2)=1,$ so for each $n$ there exists $x_n\in (0,\pi /2)$ with $f_n(x_n)=1/2,$ so $$\sup \{|f_n(x)-f(x)|:x\in [0,\pi/2]\}\ge |f_n(x_n)-f(x_n)|=|f_n(x_n)-0|=1/2.$$
A: Your explnation is correct.
First note that:
$(b)^n$ as $b$ tends to infinity is 0 if $mod(b)<1$ i.e b belongs to (-1,1) ,
 1 if b=1 and does not exists if mod(b)>1 and b=-1.
Now, $\sin x$ on 0 to $\frac{\pi}{2}$ is an increasing function and when x belongs to $[0,\frac{\pi}{2})$ sin(x) belongs to [0,1) and at $x=\frac{\pi}{2} sin(x)=1$.
Now find the pointwise limit of the function f(x) i.e lim(sin(x))^n as n tending to infinity which is f(x)={ 0 if x belongs to [0,pi/2) and 1 if x=pi/2 } now clearly the function f(x) which is the pointwise limit of f(x) is discontinuos at x= pi/2 and f_n(x)= sin^n(x) is a sequence of continuos function and there is a result that uniform limit of a sequence of continuos functions is continuos , here f(x) is discontinuos at x=1 so convergence can't be uniform.
