## keketsu Group Title I have a question about radians vs degrees when determining a limit. 2 years ago 2 years ago

1. keketsu Group Title

I'm working on the following limit: $\lim_{x \rightarrow \pi} (\sin x/-\sin x)$ I realize that this = 0, and that I need to use l'hopital's rule. My question is much simpler. This = -1 if I'm using degrees, but the correct answer of 0 (0/0) if I'm using radians. Would someone explain to me why I should be using radians?

2. Hermeezey Group Title

yes because Pi is referring to the unit circle. The Sin graph itself is in Radians, because Radians are always written with Pi

3. Hermeezey Group Title

I mean if you're dealing with Pi in trig, its a Radians problem

4. PaxPolaris Group Title

Are you sure the right answer is 0?

5. TuringTest Group Title

it can't be

6. TuringTest Group Title

also $0\neq\frac00$

7. keketsu Group Title

Hermeezey, perfect, thanks. That's the kind of thing I was looking for. So essentially, if I see a problem using pi, I need to be thinking radians, right? PaxPolaris, yeah, the problem comes with an answer. Because it comes out to 0/0, I need to find the derivative. So the ultimate answer isn't 0, but because the problem I listed comes out to 0, I need to take it further. Ah, TuringTest, I guess you're right. Infinity then?

8. TuringTest Group Title

$\frac00$is what is called an indeterminate form. That means it's just not defined, as apposed to plus or minus infinity.

9. TuringTest Group Title

In fact you can only use l'hospital's rule if by plugging in the value you are approaching you get one of the following indeterminate forms:$\frac00\text{ or }\frac{\pm\infty}0\text{ or }\frac0{\pm\infty}$if you had something that evaluated to$\pm\infty$for instance, using l'hospital is against the rules

10. keketsu Group Title

Great. You're answering questions I didn't even know I had. :) Thanks all!

11. TuringTest Group Title

anytime :D

12. keketsu Group Title

Nice, TuringTest. Thanks.

13. PaxPolaris Group Title

$\Large let\ f(x)= { \sin(x)\over -\sin(x) }$ ${\Large \therefore f(x) =-{\cancel {\sin(x)}\over \cancel {\sin(x)}}= -1} ....except\ when\ \sin(x)\ is\ 0 \ where\ it\ is\ undefined$ as you cannot divide by 0 while f(π) is undefined,$\Huge \lim_{x \rightarrow \pi}f(x)=-1$