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
Help
Have some more that I do not understand.
Limit theta approaching zero
SinTheta minus Tan theta/ sin cubed theta
2 answers
It took me a while to get that one. I cheated and used a hand calculator first, and it agrees.
5 answers