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July 31, 2014

July 31, 2014

Posted by **jane** on Monday, November 3, 2008 at 10:10pm.

Have some more that I do not understand.

Limit theta approaching zero

SinTheta minus Tan theta/ sin cubed theta

- HS Calculus -
**drwls**, Tuesday, November 4, 2008 at 9:48amRewrite 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

- HS Calculus -
**drwls**, Tuesday, November 4, 2008 at 12:05pmIt took me a while to get that one. I cheated and used a hand calculator first, and it agrees.

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