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OpenStudy (anonymous):
limits...wait a min for the equation
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OpenStudy (anonymous):
\[\lim_{x \rightarrow 0+}1/x -1/sinx\]
OpenStudy (anonymous):
infinity-infinity
OpenStudy (anonymous):
as x goes to 0+ infinity what happens to y
on the graph
that's what it's asking you
OpenStudy (anonymous):
we need to use lopetal rule
OpenStudy (anonymous):
and or 0
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OpenStudy (anonymous):
L'hospitals rule
OpenStudy (anonymous):
yep
OpenStudy (anonymous):
i am stuck while taking derivative of xsinx
OpenStudy (anonymous):
Wikipedia has everything, i'm not sure how this thing works. as well as i should. as i see it, it's what i've said above. concerning this.
OpenStudy (anonymous):
xcosx + (1)sinx
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OpenStudy (anonymous):
is the derivative of xsinx
OpenStudy (anonymous):
and the derivative of the top is 1-cosx, which as x approaches 0 = 0, cuz cosx approaches 1
OpenStudy (anonymous):
what about xcoxx?
OpenStudy (anonymous):
of course, then u need to do it again lol, and xcosx derived equals -xsinx +cosx
OpenStudy (anonymous):
so the answer is 0 right?
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OpenStudy (anonymous):
so the final thing will be sinx/(-xsinx +2cosx)
OpenStudy (anonymous):
and yes, the answer is 0
OpenStudy (anonymous):
thanks
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