`lim_(theta->(pi/2)) (1-sin(theta))/(1 + cos(2theta))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If...

`lim_(theta->(pi/2)) (1-sin(theta))/(1 + cos(2theta))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

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Chapter 4, 4.4 - Problem 15 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the limit, hence, you need to replace pi/2 for theta in limit, such that:

`lim_(theta->pi/2) (1 - sin theta)/(1 + cos 2theta) = (1 - sin(pi/2))/(1 + cos pi)`

`lim_(theta->pi/2) (1 - sin theta)/(1 + cos 2theta) = (1 - 1)/(1 + (-1)) = 0/0`

Since the limit is 0/0, then you may use l'Hospital rule:

`lim_(theta->pi/2) (1 - sin theta)/(1 + cos 2theta) = lim_(theta->pi/2) ((1 - sin theta)')/((1 + cos 2theta)')`

`lim_(theta->pi/2) ((1 - sin theta)')/((1 + cos 2theta)') = lim_(theta->pi/2) (-cos theta)/(-2sin 2theta)`

Replacing by pi/2 yields:

`lim_(theta->pi/2) (-cos theta)/(-2sin 2theta) = (-cos(pi/2))/(-2sin pi) = 0/0`

You may use again l'Hospital's rule:

`lim_(theta->pi/2) (-cos theta)/(-2sin 2theta) = lim_(theta->pi/2) ((-cos theta)')/((-2sin 2theta)')`

`lim_(theta->pi/2) ((-cos theta)')/((-2sin 2theta)')= lim_(theta->pi/2) ((sin theta)/((-4cos 2theta)`

Replacing by pi/2 yields:

`lim_(theta->pi/2) ((sin theta)/((-4cos 2theta) = (sin (pi/2))/(-4 cos pi) = 1/(-4*(-1)) = 1/4`

Hence, evaluating the limit, using twice l'Hospital's rule, yields `lim_(theta->pi/2) (1 - sin theta)/(1 + cos 2theta) = 1/4.`

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