Sine of argument with large n approximation 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.
 A: 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.
A: 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 $n \to \infty$.
A: 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! :)
A: HINT:
$$\lim_{n\to\infty}\space\frac{n\pi+\sin\left(\frac{n\pi}{2}\right)-\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$
$$\lim_{n\to\infty}\space\left(\frac{1}{2}+\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}\right)=$$
$$\lim_{n\to\infty}\space\frac{1}{2}+\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$
$$\frac{1}{2}+\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$
$$\frac{1}{2}+\frac{1}{2\pi}\lim_{n\to\infty}\space\frac{\sin\left(\frac{n\pi}{2}\right)}{n}-\frac{1}{2\pi}\lim_{n\to\infty}\space\frac{\sin\left(\frac{3\pi n}{2}\right)}{n}$$
