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Finding value of $$ \lim_{n\rightarrow \infty}\int^{1}_{0}x^{2019}\cdot \sin (nx)\,dx$$

what i try

$$I = \lim_{n\rightarrow \infty}\int^{1}_{0}x^{2019}\cdot \sin (nx)dx$$

Integration by parts

$$I=\lim_{n\rightarrow \infty}\bigg[-x^{2019}\cdot \frac{\cos (nx)}{n}\Bigg|^{1}_{0}+2019\int^{1}_{0}x^{2018}\frac{\cos(nx)}{n}dx\bigg]$$

$$I=\lim_{n\rightarrow \infty}\frac{2019}{n}\int^{1}_{0}x^{2018}\cos(nx)dx$$

How do i solve it please help me

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You are nearly done. We have $$ \vert \frac{2019}{n}\int^{1}_{0}x^{2018}\cos(nx) dx \vert \le \frac{2019}{n}\int^{1}_{0}x^{2018}dx = \frac{2019}{2019n}= \frac{1}{n} $$ Can you finish?

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You can simply apply Riemann-Lebesgue lemma.

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  • $\begingroup$ Isn't it too much for this simple problem? $\endgroup$ – who Oct 22 '19 at 14:33

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