# Limit as h->0 when [cos(pi/2+h)]/h

1. Oct 22, 2004

### UrbanXrisis

limit as h->0 when [cos(pi/2+h)]/h

when I sub in zero...I get 0/0...so I have to use L'hopital's rule
So I take the derivative of the numberator and denominator...

=[-sin(pi/2+h)*(0+1)]/1
=-sin(pi/2+h)
=-sin(pi/2+0)
= -1

is this correct?

2. Oct 22, 2004

### maverick280857

First off do you mean

$$\cos(\frac{\pi}{2} + h)$$
or
$$\cos(\frac{\pi + h}{2})$$

I think you mean the first expression

Note that $$\cos(\frac{\pi}{2} + h) = -\sin(h)$$
and the given limit equals
$$-\lim_{h->0} \frac{\sin h}{h} = -1$$

At this juncture you may use L Hospital's Theorem to reduce it to cos(h)/h but the fraction sin(h)/h goes to 1 in the limit as can be proved without L-Hospital's Theorem.