Question

Question regarding applyin L'Hopital's rule and trigonometric functions where theta is the variable: (Question below).

Answer 1

I'm supposed to find \[\lim_{\theta \rightarrow \frac{ \pi }{ 2 }}\frac{ 2\theta - \pi }{ \cos(2\pi - \theta) }\](That thing in the denominator of what's being approaches is pi, of course.) While i'm getting the right answer, Mathematica and a few other programs I use to check my work are going about it in strange ways, and I feel like i'm doing something wrong. Here's what I do:\[\lim_{\theta \rightarrow \frac{ \pi }{ 2 }}\frac{ 2\theta - \pi }{ \cos(2\pi - \theta)}= \lim_{\theta \rightarrow \frac{ \pi }{ 2 }} \frac{ 2 }{ \sin(2\pi - \theta) }\] And from there, the form is determinate; You get \[ \frac{ 2 }{ \sin(\frac{ 3\pi }{ 2 }) } = \frac{ 2 }{ -1 } = -2\] Which the book lists as correct. However, two different check methods I use are giving weird as hell results, both have worked without fail prior, and they get the same result, but they do something different. Specifically, when Wolfram derives the denominator at the beginning, it simply replaces off of the insides of it with theta. Forgive me if i'm wrong, but when you apply the chain rule to trig functions, or any for that matter, don't you derive the contents of that function, multiply the function itself by that and otherwise leave it unchanged? Why should the inside "disappear"?

tl;dr: I'm doing this at this step\[\lim_{\theta \rightarrow \frac{ \pi }{ 2 }} \frac{ 2\theta - \pi }{ \cos(2\pi - \theta) } = \frac{ 2 }{ \sin(2\pi - \theta) }\]

Answer 2

And Wolfram is doing this: \[\lim_{\theta \rightarrow \frac{ \pi }{ 2 }} \frac{ 2\theta - \pi }{ \cos(2\pi - \pi) } = -\frac{ 2 }{ \sin (\theta) }\]???

Answer 3

Wait, never mind. I see that you accounted for that.

Answer 4

(I've tried this with both taking the derivative of the denominator for Wolfram Alpha , doing the whole thing with a Wolfram Alpha widget on taking limits, a standalone L'Hopital's Rule calculator, and a Derivative calculator for the denominator, all of them do this.)

I'll post to the websites I used if anyone wants to double check. The websites go so far to display what they think you mean before it calculates it, and it matches the problem, so i'm not sure what could be going wrong.

Answer 5

Here, i'm putting in (2x-pi)/cos(2pi -x) [You can only use x as a variable in this calculator, but it should nonetheless work] as x approaches pi/2

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

And I get the right display of the function.

Answer 6

\(\cos(2\pi-x)=\cos x,\text{ since the "$2\pi-x$" part is just a shift of $2\pi$.}\)
At least, that's what it looks like WA is doing.

Answer 7

That makes perfect sense. Wow. Thanks, lol. I just started to realize that before you said it. Thanks.

Answer 8

You're welcome! WA often rewrites things without explicitly showing what it's doing, and I know it can be unhelpful at times.

Answer 9

Oh! Wait a minute. On another note, why does the theta in the trigonometric function turn positive in the Wolfram/other calculated versions? Shouldn't it still be negative?

Answer 10

@SithsAndGiggles (If you get the chance)

Answer 11

\[\begin{align*}\lim_{x\to\pi/2}\frac{2x-\pi}{\cos(2\pi-x)}&=\lim_{x\to\pi/2}\frac{2}{-\sin(2\pi-x)}\\&=\lim_{x\to\pi/2}\frac{2}{-\sin(-x)(-1)}\\
&=\lim_{x\to\pi/2}\frac{2}{\sin(- x)}\\
&=\lim_{x\to\pi/2}\frac{2}{-\sin x}\\
&=-2\end{align*}\]
Does that answer your question? (I'm not really sure what you're referring to, so I guessed.) Since sine is an odd function, you have \(\sin(-x)=-\sin x\).

Answer 12

I could understand just eliminating 2pi since it's a periodic function and that 2pi ultimately won't change the value of it, but i'm confused about why you multiply the denominator by -1 on line 2. I can understand the -sin(-x), but not where multiplying by -1 comes from.

Answer 13

The first negative sign comes from the derivative of cosine: \(\dfrac{d}{dx}\cos x=-\sin x\).

The second negative (the one next to the x inside the sine) should be clear.

The third comes from the derivative of the inside function: \(\dfrac{d}{dx}-x=-1\).

So, we have
\[\begin{align*}\frac{d}{dx}\color{blue}{\cos(\color{red}{-x})}&=\color{blue}{-\sin (-x)}\cdot\frac{d}{dx}\color{red}{(-x)}\\
&=\color{blue}{-(-\sin x)}\cdot\color{red}{(-1)}\\
&=-\sin x\end{align*}\]

Answer 14

Yup, got the first one Oh, okay. Whenever it's a value of 1 or -1, I usually assume that it's implied and never write out multiplying by that last step of the chain rule explicitly, I just change the sign. Guess I didn't catch it when it actually was. And WHOAH, since when has Openstudy's tools allowed you to color the equations???

Answer 15

I think the color-coding is part of the latex package the site uses. You'd type in \color{...}{...}, where the first argument is any color you'd like (I don't know what the range of colors is though) and the second argument is what you want colored.

Answer 16

*Awesome. Thanks.