I'm having some problem with computing (no need to prove, just compute) the limits of some trigonometric series. If you can give tips to compute, it would help me a lot in the future.
1) $$\lim_{n\to\infty} \lfloor(\sqrt n)\rfloor \sin\frac{1}{n} $$ 2) $$\lim_{n\to\infty} \sin\frac{\pi n}{2}*\cos\left(\sin \frac{1}{n}\right) $$ 3) Let the range of a be (0,$\pi$). Prove that: $$\lim_{n\to\infty} ((2\pi n+a)\sin(2\pi n+a)-(2\pi n\sin(2\pi n))= \infty $$
Problems I've had:
The problems I'm having are mostly with the floor limit with the sqrt, and the limits of trigonometric functions. I also tend to miss side limits. Is there a way to make sure after calculating the limits that I did not miss one?
I'd very appreciate if you could also provide me tips or ideas for an efficient way of solving those limits.
(no taylor or l'hopital please)
For example: using $\;\lfloor x\rfloor=x-\{x\}\;,\;\;\{x\}:=\;$ the fractional part of $\;x\;$ , we get:
$$\lfloor n\rfloor\sin\frac1n=\left(n-\{n\}\right)\sin\frac1n=\frac{\sin\frac1n}{\frac1n}-\{n\}\sin\frac1n\xrightarrow[n\to\infty]{}1-0=1$$
Try now the other ones.