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keketsu
I have a question about radians vs degrees when determining a limit.
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?
yes because Pi is referring to the unit circle. The Sin graph itself is in Radians, because Radians are always written with Pi
I mean if you're dealing with Pi in trig, its a Radians problem
Are you sure the right answer is 0?
also \[0\neq\frac00\]
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?
\[\frac00\]is what is called an indeterminate form. That means it's just not defined, as apposed to plus or minus infinity.
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
Great. You're answering questions I didn't even know I had. :) Thanks all!
Nice, TuringTest. Thanks.
\[\Large let\ f(x)= { \sin(x)\over -\sin(x) }\] \[{\Large \therefore f(x) =-{\cancel {\sin(x)}\over \cancel {\sin(x)}}= -1} ....except\ when\ \sin(x)\ is\ 0 \ where\ it\ is\ undefined\] as you cannot divide by 0 while f(π) is undefined,\[\Huge \lim_{x \rightarrow \pi}f(x)=-1\]