HS Calculus
posted by jane .
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Have some more that I do not understand.
Limit theta approaching zero
SinTheta minus Tan theta/ sin cubed theta

Rewrite as [sin x(1  1/cosx)]/sin^3x
= [1  (1/cosx)]/[1  cos^2x]
Now use L'Hopital's rule and take the ratio of the derivatives of numerator and denominator. That lets you get rid of the 1's.
You are left with
Lim [tanx secx/(2cosx sinx)]
= Lim 1/(2 cos^3x)
Which is 1/2 
It took me a while to get that one. I cheated and used a hand calculator first, and it agrees.
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