# What are the steps to solve this limit $\lim _{x\to 0} \frac{\cos(x+\pi/2)}x$?

$\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi }{2}\right)}{x}\right)\:$

• do you know the rules of L'Hospital? Oct 31, 2015 at 19:56
• A very good first step would be $\cos(x+\pi/2)=-\sin x$... Oct 31, 2015 at 19:57
• @Dr.SonnhardGraubner No I don't Oct 31, 2015 at 20:00
• @user283144 It's good that you don't. It's a lousy way to do this.
– zhw.
Oct 31, 2015 at 21:11

$\cos(x+\frac{\pi}{2})=-\sin(x)$

So $\displaystyle\lim_{x \to 0} \frac{-\sin(x)}{x}=-1$

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.

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.

$$\lim _{x\to 0}\left(\frac{\cos\left(x+\frac{\pi}{2}\right)}{x}\right)=$$ $$\lim _{x\to 0}\left(-\frac{\sin(x)}{x}\right)=$$ $$-\left(\lim _{x\to 0}\frac{\sin(x)}{x}\right)=$$ $$-\left(\lim _{x\to 0}\frac{\frac{\text{d}}{\text{d}x}\sin(x)}{\frac{\text{d}}{\text{d}x}x}\right)=$$ $$-\left(\lim _{x\to 0}\frac{\cos(x)}{1}\right)=$$ $$-\left(\lim _{x\to 0}\cos(x)\right)=$$ $$-\left(\cos(0)\right)=$$ $$-\left(1\right)=-1$$ 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 $$\lim_{x\to 0}\frac{\cos(x+\frac{\pi}{2})}{x} = \lim_{x\to 0}\frac{-\sin x}{x} = \dots$$