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Limits and derivatives

  1. Sep 18, 2005 #1
    i have a test tomorrow but i have no idea how to do these probelms...
    1) lim x->0 sin2x/sin5x

    2) write an equation for the stright line through (1,0) that is tangent to the graph of y= x + 1/x....

    thanks in advance
     
  2. jcsd
  3. Sep 18, 2005 #2

    HallsofIvy

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    Is there a reason for posting this under "differential equations"?? I'm going to move it to "Homework, k-12". I think it will get a better response there.

    1) What if this were [itex]lim_{y,z->0}\frac{sin y}{y}\frac{z}{sin z}[/itex]? Could you do it then? Can you think of a way to change it into that form?

    2) One definition of "derivative" is that it is the "slope of the tangent line". Would knowing the slope of the line help?
     
  4. Sep 18, 2005 #3
    sorry, i was in a rush so i posted it in the wrong section :biggrin:
    1) i tried (2x/2x) (sin2x)/sin(5x) -> (2x/sin5x) (sin2x)(2x)..now i dont know how to get rid of the (2x/sin5x)

    2) hmm...the derivative i got was (X^2-1)/x^2. do i just plug in a random number to get a slope then use point sliope formula to make it include (1,0)?
     
  5. Sep 19, 2005 #4

    Diane_

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    1) Note that you have the indeterminate form 0/0. Does this suggest anybody's Rule?

    2) You don't use a random point. You have to find a point (x, y) such that the slope of the line (which is constant, of course) equals the slope of the curve at that point. Note that the line will also pass through that point. Think simultaneous equations.
     
  6. Sep 19, 2005 #5

    HallsofIvy

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    L'Hopital's rule will work (easily) but I would consider it overkill! You wouldn't want to do this the easy way, would you?

    MercuryRising: You changed [tex]\frac{sin 2x}{sin 5x}[/tex] to [tex]\frac{sin 2x}{2x}\frac{5x}{sin 5x}[/tex] by multiplying the numerator by 5x and denominator by 2x. Of course you can't do that! But you can multiply both numerator and denominator by the same thing. That gives [tex]\frac{sin 2x}{2x}\frac{5x}{sin 5x}\frac{2x}{5x}[/tex]. Can you find the limit of that?
     
  7. Sep 19, 2005 #6

    Diane_

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    As a matter of fact, yes. :smile:
     
  8. Sep 26, 2005 #7

    Tom Mattson

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    Maybe L'Anybody's Rule would habe been a better hint. :rofl:
     
  9. Sep 26, 2005 #8

    HallsofIvy

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    Do you have a cold, Tom?
     
  10. Sep 26, 2005 #9

    Tom Mattson

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    No, I do not habe a cold. :tongue2:
     
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