HELP - LIMITS - URGENT AP TEST TOMORROW

[InterserveVB]

Can anyone explain how to do this problem? The answer is supposed to be 1.

 

[InterserveVB]

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.

 

[soroban]

Hello, InterserveVB!

\(\displaystyle \L\lim_{h\to0}\frac{\cos\left(\frac{3\pi}{2}\,+\,h\right)\,-\,\cos\left(\frac{3\pi}{2}\right)}{h}\)

The numerator is: \(\displaystyle \,\cos\left(\frac{3\pi}{2}\right)\cos(h) \,- \,\sin\left(\frac{3\pi}{2}\right)\sin(h) \,-\,\cos\left(\frac{3\pi}{2}\right)\)

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

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