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Find the limit. Use l'Hospital's Rule where appro…

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Problem 17 Medium Difficulty

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.

$ \displaystyle \lim_{\theta \to \pi /2} \frac{1 - \sin \theta}{1 + \cos 2\theta} $

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Video Transcript

Hello. So here we consider the limit as theta approaches pi over two of one minus sine of theta divided by one plus co sign of tooth data. So so to um so we can use lava tiles rule. We have to first make sure we have an indeterminate form um that we can use the total Israel on. So we're gonna go ahead and just do direct substitution first. So if we do that we have one minus sine of pi over two. Um That's gonna be one minus one. Sine of pi over two is once we have one minus one and then divided by one plus. Co sign of two times pi over two. Well two times pi over two is just pie. So we have one plus co sign of pi or co sign a pie is negative one. So if we have one plus a negative one which gives us well zero divided by zero. So yes, we're in the form here 0/0. So lumpy tiles rule can be applied. So we then go ahead and take the derivative of the top and divided by the derivative of the bottom. So we then have the limit as data approaches, pi over two of while the derivative of one minus sine of theta is gonna be negative um negative coastline of data. So we have negative co sign of data and then divided by the derivative of one plus co sign of two data is going to be well negative to sign of tooth data by the train will so we have divided by negative two sign of two data and then going ahead and taking the limit here. Well putting in pi over two. We have negative co sign of pi over two. That's going to be just um or negative zero. So we get negative. Well zero divided by um Hi over to in the bottom. What do we have? We would have negative two times sine of two times pi over two. That's just negative two times sine of pi which is negative two times zero. So again we get zero over zero. So again we have an indeterminate form that we can go ahead and use low petals rule again. So again what we have here we have the limit as fattah approaches pi over two. Now taking the derivative of this thing. So we take the derivative of negative co sign of data. So that is going to give us um while negative negative center. That is a positive sign of feta. And then divided by the derivative of uh negative to sign of uh to to to that is going to give us um coastline of tooth data. So co sign of tooth data. Now in putting in pi over two. We get here well we get a negative 1/4 times sine of pi over two divided by co sign of two times pi over two. So we get a um negative 1/4 and then taking the limit here we get well times sine of pi over two divided by co sign of pie. So this is going to give us negative 1/4 times while one over negative one, which is just a negative 1/4 times negative one, which gives us 1/4. So therefore the limit here evaluates to 1/4. Take care.

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Calculus: Early Transcendentals

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