spacewater
Junior Member
- Joined
- Jul 10, 2009
- Messages
- 67
problem
\(\displaystyle \frac{lim}{?x->0} \frac{sin[(\pi/6)+?x]-( \frac{1}{2})}{?x}\) (Find the limit when ?x approaches 0)
Steps
\(\displaystyle \frac{sin(\pi/6)cos?x+cos(\pi/6)sin?x-(1/2)}{?x}\)
\(\displaystyle \frac {1/2cos?x+\sqrt3/2sin?x-1/2}{?x}\)
this is where I got stuck... 1/2 * cos(0) +\(\displaystyle \sqrt3\)/2*sin(0) - 1/2 = 1/2-1/2 = 0
Can someone point out where I went wrong on this equation please?
\(\displaystyle \frac{lim}{?x->0} \frac{sin[(\pi/6)+?x]-( \frac{1}{2})}{?x}\) (Find the limit when ?x approaches 0)
Steps
\(\displaystyle \frac{sin(\pi/6)cos?x+cos(\pi/6)sin?x-(1/2)}{?x}\)
\(\displaystyle \frac {1/2cos?x+\sqrt3/2sin?x-1/2}{?x}\)
this is where I got stuck... 1/2 * cos(0) +\(\displaystyle \sqrt3\)/2*sin(0) - 1/2 = 1/2-1/2 = 0
Can someone point out where I went wrong on this equation please?