HELP - LIMITS - URGENT AP TEST TOMORROW

ok, never mind


Cos (3pi / 2) = - sin (3pi/2) = 1
I was doing the D of the inside and it was causing it to 0 out.
 
Hello, InterserveVB!

\(\displaystyle \L\lim_{h\to0}\frac{\cos\left(\frac{3\pi}{2}\,+\,h\right)\,-\,\cos\left(\frac{3\pi}{2}\right)}{h}\)
Click to expand...
The numerator is:  cos⁡(3π2)cos⁡(h) − sin⁡(3π2)sin⁡(h) − cos⁡(3π2)\displaystyle \,\cos\left(\frac{3\pi}{2}\right)\cos(h) \,- \,\sin\left(\frac{3\pi}{2}\right)\sin(h) \,-\,\cos\left(\frac{3\pi}{2}\right)

. . . . . . . . . . . . =  0⋅cos⁡(h) − (−1)sin⁡(h) − 0  =  sin⁡h\displaystyle = \;0\cdot\cos(h)\,-\,(-1)\sin(h) \,-\,0\;=\;\sin h

Then we have: \(\displaystyle \L\,\lim_{h\to0} \frac{\sin h}{h}\;=\;1\)
 
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