Finding the limit of a sequence defined as an integral

Question

I have two sequences defined in terms of an integral: $$a_n=\int_{-n\pi}^{n\pi}\sin x\,dx$$ $$b_n=\int_{-2n\pi}^{(2n+1)\pi}\sin x\,dx$$ And I am tasked with finding the limit as $n$ approaches infinity. I am not really sure how to achieve this. I have looked at another question on this site that is relatively similar and they used substitution however I have no idea what I should let $u$ equal to make this work out nicely and I did not see this explained.

Answer 1

First evaluate the integrals: $$a_n=\int_{-n\pi}^{n\pi}\sin x\,dx=[-\cos x]_{-n\pi}^{n\pi}=0$$ $$b_n=\int_{-2n\pi}^{(2n+1)\pi}\sin x\,dx=[-\cos x]_{-2n\pi}^{(2n+1)\pi}=2$$ Both $a_n$ and $b_n$ are constant sequences, so their limits as $n\to\infty$ are 0 and 2 respectively.

Answer 2

You can actually compute the integral and express $a_n$ and $b_n$ without using the integral. Note that \begin{align*} a_n=\int_{-n\pi}^{n\pi}\sin(x)\ dx&=\left.-\cos(x)\right|_{-n\pi}^{n\pi}\\&=-\cos(n\pi)+\cos(-n\pi)\\&=-\cos(n\pi)+\cos(n\pi)\\ &=0, \end{align*} so $\lim\limits_{n\rightarrow\infty} a_n = 0$.

Again, using the fact that $\cos$ is a periodic function with period $2\pi$, note that \begin{align*} b_n=\int_{-2n\pi}^{\pi+2n\pi}\sin(x)\ dx&=\left.-\cos(x)\right|_{-2n\pi}^{\pi+2n\pi}\\&=-\cos(\pi+2n\pi)+\cos(-2n\pi)\\&=-\cos(\pi)+\cos (0)\\ &=2 \end{align*} so $\lim\limits_{n\rightarrow\infty} b_n = 2$.