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

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?

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 $$n$$].
If the interval includes pi, then there is no pointwise convergence. Otherwise there is and to prove that it is not uniform show that $$\forall n \exists x \in (0,π)\ f_n(x)<1/2$$
• For $x=\pi$, $f_n(x)=0$ for all $n$, so pointwise limit does exist. – Kabo Murphy Sep 17 at 12:33