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keketsu

  • 4 years ago

I have a question about radians vs degrees when determining a limit.

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  1. keketsu
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    I mean if you're dealing with Pi in trig, its a Radians problem

  4. PaxPolaris
    • 4 years ago
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    Are you sure the right answer is 0?

  5. TuringTest
    • 4 years ago
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    it can't be

  6. TuringTest
    • 4 years ago
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    also \[0\neq\frac00\]

  7. keketsu
    • 4 years ago
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    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
    • 4 years ago
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    \[\frac00\]is what is called an indeterminate form. That means it's just not defined, as apposed to plus or minus infinity.

  9. TuringTest
    • 4 years ago
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    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
    • 4 years ago
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    Great. You're answering questions I didn't even know I had. :) Thanks all!

  11. TuringTest
    • 4 years ago
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    anytime :D

  12. keketsu
    • 4 years ago
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    Nice, TuringTest. Thanks.

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

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