How do you find the limit of (1-sin(theta))/(1+cos(2theta))1−sin(θ)1+cos(2θ) as theta approaches pi/2?
2 Answers
If you are trying to do this without l'Hospital, see below.
Explanation:
We'll use some trigonometry to rewrite the expression to get a determinate form.
So let's multiply by
At the same time we'll use a double angle formula for
= 1-2sin^2 theta=1−2sin2θ " " might be helpful, but we're going to getcos thetacosθ on top, so let's go with
Try it! If it doesn't work, try something else. Do not just give up if your first attempt doesn't work.
= (1-sin^2 theta)/(2cos^2 theta (1+sin theta))=1−sin2θ2cos2θ(1+sinθ)
= (cos^2 theta) / (2cos^2 theta (1+sin theta))=cos2θ2cos2θ(1+sinθ)
= 1/(2(1+sin theta)=12(1+sinθ)
Taking the limit as
Alternative Solution using
= (1-sin theta)/(2(1-sin^2 theta))=1−sinθ2(1−sin2θ)
Now
= (1-sin theta)/(2(1+sin theta)(1-sin theta)=1−sinθ2(1+sinθ)(1−sinθ)
= 1/(2(1+sin theta)=12(1+sinθ)
Taking the limit as
Explanation:
We know that,
Reqd. Limit
Let
obtained!
Enjoy maths.!.