What are the steps to solve this limit $\lim _{x\to 0} \frac{\cos(x+\pi/2)}x$? $\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi }{2}\right)}{x}\right)\:$
 A: Hint. You may observe that
$$
\cos\left(x+\frac{\pi }{2}\right)=-\sin x
$$ then your limit is equal to
$$
-\lim _{x\to 0}\left(\frac{\sin x}{x}\right)=?
$$ which is easier to obtain.
A: Another suggestion which may work more generally (although the above solutions are better in this case) is to apply L'Hôpital's Rule for an indeterminate form $\frac{0}{0}$. That is, in this case, the limit of the derivatives of the numerator and denominator is equal to the limit of the original expression:
$$ \lim_{x\to 0}\frac{\cos(x+\frac{\pi}{2})}{x} = \lim_{x\to 0}\frac{-\sin(x+\frac{\pi}{2})}{1}=\lim_{x\to0}-\sin(x+\frac{\pi}{2})=-\sin(\frac{\pi}{2})=-1.$$
The same result as in the other answers. Use this technique any time you have an indeterminate form.
A: $\cos(x+\frac{\pi}{2})=-\sin(x)$
So $\displaystyle\lim_{x \to 0} \frac{-\sin(x)}{x}=-1$
A: $$\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi}{2}\right)}{x}\right)=$$
$$\lim _{x\to 0}\left(-\frac{\sin(x)}{x}\right)=$$
$$-\left(\lim _{x\to 0}\frac{\sin(x)}{x}\right)=$$
$$-\left(\lim _{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin(x)}{\frac{\text{d}}{\text{d}x}x}\right)=$$
$$-\left(\lim _{x\to 0}\frac{\cos(x)}{1}\right)=$$
$$-\left(\lim _{x\to 0}\cos(x)\right)=$$
$$-\left(\cos(0)\right)=$$
$$-\left(1\right)=-1$$
A: 
For acute angles $\alpha$ and $\beta$, use the law of cosines to show that
$$\cos(\alpha+\beta)=\cos\alpha\cos\beta - \sin\alpha\sin\beta$$
Then, use this identity to show that
$$\cos\bigg(x+\frac{\pi}{2}\bigg) = -\sin x$$
and, finally evaluate the limit
$$\lim_{x\to 0}\frac{\cos(x+\frac{\pi}{2})}{x} = \lim_{x\to 0}\frac{-\sin x}{x} = \dots$$
