# Limit without using l'Hopital

How can I find $$\lim_{n \to \infty}{n \cos(\pi/2 + 1/n)}$$ without using L'Hôpital's rule?

Using L'Hopital's rule I can find the answer to be -1, but without it I don't know where to start.

Hint: $$\cos\left(\frac{\pi}{2}+\frac{1}{n}\right)=\sin(?).$$
HINT: $$\cos\left(\frac{\pi}2+\frac1n\right)=\cos\frac{\pi}2\cos\frac1n-\sin\frac{\pi}2\sin\frac1n\;;$$ what are $\cos\frac{\pi}2$ and $\sin\frac{\pi}2$?
HINT $$\cos\left(\dfrac{\pi}2+y \right) = - \sin \left(y \right)$$ and $$\lim_{x \to 0}\dfrac{\sin(x)}{x} = 1$$
Put $m=\frac{1}{n}$, then
$$\lim_{n \to \infty}{n \cos(\pi/2 + 1/n)} = \lim_{m \to 0}\frac{\cos(m+\pi/2 )}{m} =\dots$$