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OpenStudy (anonymous):
Using L'Hopital's Rule, what is the limit of: (sin2x)/(sin7x) As x approaches 0?
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OpenStudy (p0sitr0n):
derive up and bottom part sin2x/sin7x=2cosx/7cosx = 2/7 as x = 0
OpenStudy (anonymous):
My online HW says this is right. But isn't the derivative of sin2x equal to 2cos2x?
OpenStudy (anonymous):
Wait I got it, cos(0) is equal to 1, duh
OpenStudy (anonymous):
THE LIMIT DOES NOT EXIST!!!
OpenStudy (anonymous):
sorry mean girls quote
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OpenStudy (anonymous):
lol get it...
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