Check whether the following sequence of functions are uniformly convergent or not? $$ f_n(x)=(\sin(x))^n,\qquad x \in [0,π/2) $$
Limiting function is continuous. So , I could not do anything with that. I tried to use M-test, it is not working. Please help me.
Short answer: no, there is no uniform convergence. To see how to prove this, let us come up with some sequence $(x_n)_n$ such that $f_n(x_n)$ does not converge to $0$. Below, I describe in detail how to find such a sequence.
Since $\sin x \to 1$ when $x\to \frac{\pi}{2}$, intuitively the issue should be "around $\frac{\pi}{2}$." Thus, we want to find $x_n$ of the form $\frac{\pi}{2}-o(1)$, for the "right" $o(1)$.
Let $(\varepsilon_n)_n$ be a sequence converging to $0$ that we will pick later, according to our needs, and consider $$x_n \stackrel{\rm def}{=} \frac{\pi}{2}-\varepsilon_n$$ for $n\geq 1$. Then $$\begin{align} f_n(x_n) &= \sin(\frac{\pi}{2}-\varepsilon_n)^n = (\cos \varepsilon_n)^n = \left(1-\frac{\varepsilon_n^2}{2}+o(\varepsilon_n^2)\right)^n \\&= \exp\left(n\ln\left(1-\frac{\varepsilon_n^2}{2}+o(\varepsilon_n^2)\right)\right)\\& = \exp\left(-\frac{n\varepsilon_n^2}{2}+o(n\varepsilon_n^2)\right) \end{align}$$ using the Taylor expansions of $\cos u$ and $\ln(1+u)$ at $0$ (and the assumption that $\varepsilon_n \xrightarrow[n\to\infty]{}0$).
This means that, choosing ${n\varepsilon_n^2}={2}$, i.e. $$ \varepsilon_n \stackrel{\rm def}{=} \sqrt{\frac{2}{n}} $$ we get, for $x_n=\frac{\pi}{2}-\sqrt{\frac{2}{n}}$, that $$\begin{align} f_n(x_n) &= \exp\left(-1+o(1)\right) \xrightarrow[n\to\infty]{} \frac{1}{e} \neq 0. \end{align}$$ Thus, we do not have uniform convergence, as $$ \sup_{x\in(0,\frac{\pi}{2})} \lvert f_n(x)\rvert \geq \lvert f_n(x_n)\rvert \xrightarrow[n\to\infty]{} \frac{1}{e} > 0. $$