Sine of argument with large n approximation

Question

I have worked an integral and reduced the integral to $$\frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}$$

I want to show that for $$n\rightarrow \infty$$ the above equation reduces to $$\frac{1}{2}$$

Evidently, this means the $2$ sine functions must cancel each other. But what is a good way to do this? Large $n$ results in sine toggling between $-1$ and $1$. Notice that the either sine function will have a sign opposite to the other.

Answer 1

The two $\sin$ functions needs not cancel each other. We have $$\sin(nx) - \sin(3nx) \in [-2,2]$$ Hence, we have $$\dfrac{n\pi-2}{2n\pi} \leq \dfrac{n\pi + \sin(nx) - \sin(3nx)}{2n\pi} \leq \dfrac{n\pi+2}{2n\pi} $$ Now take $n \to \infty$.

Answer 2

First notice that

$$\begin{align} \sin(\frac{3n\pi}{2})&=\sin(\frac{n\pi}{2}+n\pi) \\ &=\sin(\frac{n\pi}{2})\cos(n\pi)+\cos(\frac{n\pi}{2})\sin(n\pi) \\ &=\sin(\frac{n\pi}{2})\cos(n\pi)+0 \\ &=(-1)^n\sin(\frac{n\pi}{2}) \end{align}$$

and hence

$$\sin(\frac{n\pi}{2})-\sin(\frac{3n\pi}{2})=(1-(-1)^n)\sin(\frac{n\pi}{2})$$

The above expression is $0$ or $2$ depending on $n$ is even or odd. Now, you can find a bound easily

$$0 \le (1-(-1)^n)\sin(\frac{n\pi}{2}) \le 2$$

And finally

$$\frac{1}{2} \le \frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}\le \ \frac{1}{2} + \frac{1}{ n \pi}$$

Now, you can guess what happens next! :)

Answer 3

Divide numerator and denominator by $n$: $$ \frac{n\pi+\sin\left(\frac{n \pi}2\right)-\sin\left(\frac{3\pi n}2\right )}{2n\pi} = \frac{\pi+\frac{\sin\left(\frac{n\pi}2\right)}n-\frac{\sin\left(\frac{3\pi n}2\right )}n}{2\pi}\to\frac{\pi+0+0}{2\pi} = \frac12 $$ because $\sin$ is a bounded function.

Answer 4

HINT:

$$\lim_{n\to\infty}\space\frac{n\pi+\sin\left(\frac{n\pi}{2}\right)-\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$ $$\lim_{n\to\infty}\space\left(\frac{1}{2}+\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}\right)=$$ $$\lim_{n\to\infty}\space\frac{1}{2}+\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$ $$\frac{1}{2}+\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$ $$\frac{1}{2}+\frac{1}{2\pi}\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{n}-\frac{1}{2\pi}\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{n}$$