# Math Help - limit problem

1. ## limit problem

I have no idea how to solve this..can somebody give me some help? thanks

2. $\cos \left( {\frac{\pi }{2} + h} \right) = - \sin (h)$

3. Originally Posted by jst706
I have no idea how to solve this..can somebody give me some help? thanks
Expand $\cos(\pi/2+h)$ using the multiple angle formula:

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

........... $=\frac{\cos(\pi/2)(\cos(h)-1)-\sin(\pi/2)\sin(h)}{h}
$

........... $=\frac{\cos(\pi/2)(\cos(h)-1)}{h}-\frac{\sin(\pi/2)\sin(h)}{h}
$

The limit of the first term above is $0$, and the second is $-\sin(\pi/2)$

RonL