Uniform convergence of $\left(\frac{\sin x}{x}\right)^{\frac1n}$

Question

The sequence $\displaystyle{f_n(x)= \left(\frac{\sin x}{x}\right)^{\frac{1}{n}},\, x\in [0,π] .}$ Find whether it converges pointwise, uniformly or not?

Since $\displaystyle{\lim_{n \to \infty} \left(\frac{\sin x}{x}\right)^{\frac{1}{n}} = 1 \,\forall x \in [0,π]}$, it is pointwise convergent. I know it's not uniform convergent but how to approach for that?

Answer 1

For $x=\pi$ the pointwise limit is not $1$, it is $0$. This also proves that the convergence is not uniform because the pointwise limit is not a continuous function.

[I am interpreting $f_n(0)$ as $1$ for all $n$].

Answer 2

If the interval includes pi, then there is no pointwise convergence. Otherwise there is and to prove that it is not uniform show that $$\forall n \exists x \in (0,π)\ f_n(x)<1/2$$