Question

Another L'Hopital's Rule problem regarding trigonometric functions and angles.

(Problem below).

Answer 1

I'm supposed to find\[\lim_{\theta \rightarrow \pi/2}\frac{ 1 - \sin(\theta) }{ 1+\cos(2\theta) }\]This is another weird one where I get the right answer, which i'm sure of, but if I check it with Wolfram Alpha or other tools, some really odd method is undertaken. I've had other problems similar but not quite like this regarding shifts of 2pi where things became negligible by that, but I haven't done anything like this with double angles. What i'm doing is this:\[\lim_{\theta \rightarrow \pi/2}\frac{ 1-\sin(\theta) }{ 1+\cos(2\theta)} = \lim_{\theta \rightarrow \pi/2}\frac{ \cos (\theta) }{ 2\sin(2\theta)}= \lim_{\theta \rightarrow \pi/2}\frac{ -\sin(\theta) }{ 4\cos(2\theta) }=\frac{ -1 }{ -4 }=\frac{ 1 }{ 4 }\]
However, when I do this with Wolfram Alpha, it returns the same answer but does this:\[\lim_{\theta \rightarrow \pi/2}\frac{ 1-\sin(\theta) }{ 1+\cos(2\theta)} =\lim_{\theta \rightarrow \pi/2} \frac{ 1 }{ 4\sin(\theta) }\]I just don't understand at all how WA got here. Any help would be appreciated on explaining this. Thanks.

Answer 2

You can probably get their result using some manipulation (prob with the double angle formula on cos2theta).

Answer 3

Yeah, and that's what i'm banking on it being, but the weirder thing is that you know how WA explicitly states steps? It says that that should be the first result of applying L'Hopital's Rule to the original limit. I'll link to the WA widget i'm using and explicitly show my input:

Answer 4

WA sometimes takes odd methods to get to solutions, though.

Answer 5

Yeah. I dunno, I might just ask my Calc professor when I get the chance. I'm just curious how it did this if it isn't "lying" and really only applied L'Hopital's Rule.

Input: The limit of "(1-sin(x))/(1+cos(2x))" asx approaches "pi/2" (copied and pasted directly)

http://www.wolframalpha.com/widgets/view.jsp?id=265eceb6d4d961057f1b483a558e2885

The explicit steps should be outlined if you click to expand it, I might be missing something.

Answer 6

(Substituted x for theta, you can only use x as a variable in this widget)

Answer 7

lol, looking at that I thought "the f---?". I'm gonna try to find how they got from l'hopitals rule to that solution.

Answer 8

Yeah. I just, I dunno, lol. I just, http://i1152.photobucket.com/albums/p495/darkfire814/Gifs/f45.gif

Answer 9

Oh I see. They only applied the rule once, then used sin2x = 2sinxcosx.

Answer 10

Okay, so it was based on trig identities. Nevermind, i'm not crazy. Jesus, that's one thing they really need to fix; whenever a limit is calculated, it never fails to never mention applying an identity, in this situation and others.

Answer 11

\[\large \lim_{\theta \rightarrow \pi/2}\frac{ \cos (\theta) }{ 2\sin(2\theta)} = \lim_{\theta \rightarrow \pi/2}\frac{ \cos (\theta) }{ 2(2\sin \theta \cos \theta)} = \lim_{\theta \rightarrow \pi/2}\frac{ 1 }{ 4\sin \theta} \]

Answer 12

Yup, makes sense. Got it.