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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?

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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 $n$].

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

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  • $\begingroup$ For $x=\pi$, $ f_n(x)=0$ for all $n$, so pointwise limit does exist. $\endgroup$ – Kabo Murphy Sep 17 at 12:33

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