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Find the limit of (2y - 180°)/cos y as y-> 90°. How to find the answer

  1. Aug 1, 2009 #1
    please give a hint on how to solve this.

    the limit of (2y - 180 degrees)/(cos y) 'as y approaches 90 degrees.'

    the answer here is -2.

    I need a 'hint' for me to arrive at -2.

    thank you
     
    Last edited: Aug 1, 2009
  2. jcsd
  3. Aug 1, 2009 #2

    Dick

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    Re: limits

    Try substituting u=y-90 degrees.
     
  4. Aug 1, 2009 #3
    Re: limits

    what "u"?

    where should I substitute it?

    thank you
     
  5. Aug 1, 2009 #4

    jgens

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    Re: limits

    Here, let y = u + 90 and substitute that in whenever you have a y in your equation.
     
  6. Aug 1, 2009 #5
    Re: limits

    thank you

    but why should I let y be equal to u+90?

    since y approaches 90 degrees, to evaluate the limit, what I have learned is to just substitute y= 90 degrees.

    so how is that?
     
  7. Aug 1, 2009 #6
    Re: limits

    Do you know how to differentiate functions? If so, you could use L'Hôpital's rule here.
     
  8. Aug 1, 2009 #7
    Re: limits

    On a different note, if you substitute in [tex]y=u+90[/tex] and [tex]u=y-90[/tex], you get rid of dividing by 0. You'll have to find the "[tex]u[/tex] equivalent" of 90 because all of your variables will be with respect to [tex]u[/tex], so approaching the [tex]y[/tex] value just wouldn't make sense.

    EDIT: It took me a while to figure out what Dick and Jgens meant with the substitutions. I've forgotten how to take weird limits like this without using l'Hopital's rule, lol.
     
  9. Aug 1, 2009 #8

    Dick

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    Re: limits

    The point of the substitution is to turn part of the limit into something you know. Like sin(u)/u. cos(u+90 degrees) is -sin(u), isn't it?
     
  10. Aug 2, 2009 #9

    VietDao29

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    Re: limits

    Maybe I'm missing something here, but I don't really get it. Are you sure it's not:

    [tex]\lim_{y \rightarrow \frac{\pi}{2}} \left( \frac{2y - \pi}{\cos y} \right)[/tex]?

    The well-known limit:

    [tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex] is only true when x is in radians. And this is one of the reasons, that we use radians to do calculus work.
     
  11. Aug 2, 2009 #10

    Dick

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    Re: limits

    Now that's not true. lim x->0 of sin(kx)/(kx)=1. That means lim sin(x)/x=1 is true in any units. The reason for working in radians is so that d/dx(sin(x))=cos(x). And not cos(x) times some funny unit thing.
     
  12. Aug 3, 2009 #11

    VietDao29

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    Re: limits

    No, this is not true.

    You can just try to plug some x near 0, in Degree Mode, and you can see the different.

    In fact, the reason for d/dx(sin(x)) (x in degrees) does not equal cos(x) is because of this limit. This limit is different when x is in different modes.

    Since we have:
    [tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1[/tex] where x is in radians.

    So, to find the limit:
    [tex]\lim_{x \rightarrow 0} \frac{\sin x}{x}[/tex], where x is in degrees, we have to change it back to radians. The difference when changing from degrees to radians only lies in the numerator, since [tex]\sin(x_0 \mbox{ [in Rad]} ) \neq \sin(x_0 \mbox{ [in Deg]})[/tex]

    So:

    [tex]\lim_{x \rightarrow 0} \frac{\sin x}{x} = \lim_{x \rightarrow 0} \frac{\sin \frac{\pi}{180} x}{x}[/tex]
    [tex]= \lim_{x \rightarrow 0} \frac{\sin \frac{\pi}{180} x}{\frac{\pi}{180}x} \times \frac{\pi}{180} = \frac{\pi}{180}[/tex].
     
  13. Aug 3, 2009 #12

    Dick

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    Re: limits

    You are absolutely right. Sorry, you only convert the numerator sin(x) to radians, not the denominator.
     
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