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

Question

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

Answer 1

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?

Answer 2

You can simply apply Riemann-Lebesgue lemma.