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Trig derivative

  1. May 27, 2006 #1
    What is the derivative of [tex]cos\frac{1}{x}[/tex]?

    Also, would [tex]cos\frac{1}{x}[/tex] equal 0 or infinity?
     
    Last edited: May 27, 2006
  2. jcsd
  3. May 27, 2006 #2
    Please post your attempt(s) in finding it.

    The cosine function can have values only within the interval [-1,1], and has the value 0 for certain values of x. (an example in this case: [tex] x = \frac{2}{\pi} [/tex])
     
  4. May 27, 2006 #3
    if you put: x=1/2pi*n
    you get cos' (2pi*n)=(-2pi)sin(1/x)
     
  5. May 27, 2006 #4
    I attempted to use the chain rule and I got [tex]-\frac{sin}{x^2}\left(\frac{1}{x}\right)[/tex]

    I'm trying to solve a limit where [tex]\lim_{x \rightarrow 0} \left[3x^5cos\left(\frac{1}{x}\right)\right][/tex].

    So I'm using Indeterminate forms and L'Hopital's rule to try and solve it.
     
    Last edited: May 27, 2006
  6. May 27, 2006 #5
    my mistake it should be:
    (1/x^2)sin(1/x)

    edit:
    about the limit, i don't think it exists, cause as i said it's cos y wher y approaches infinity, which is not defined.
     
    Last edited: May 27, 2006
  7. May 27, 2006 #6
    [tex]\frac{1}{x^2}sin(\frac{1}{x}) [/tex] is correct. The two minus's cancel.

    EDIT: Stupid me...getting this Latex all messed up! Sorry. Rewrite that as
    [tex]\lim_{x \rightarrow 0} \left[3x^4\frac{cos\left{(\frac{1}{x}\right)\right}{\frac{1}{x}}][/tex]
     
    Last edited: May 27, 2006
  8. May 27, 2006 #7
    Alright, thanks for the first part.

    So [tex]\frac{1}{x}, x = 0[/tex], would be infinity and so, [tex]sin(\infty)[/tex] and [tex]cos(\infty)[/tex] would be indeterminate?
     
  9. May 27, 2006 #8
    I'm really sorry. As I said I'm getting both the logic and the LATEX wrong.:redface: I'll let someone better tackle this while I go hide my face from the public.
     
  10. May 27, 2006 #9

    VietDao29

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    Homework Helper

    Err, sorry to interrupt you guys, but why can't it (i.e the limit) exist? :confused:
     
  11. May 27, 2006 #10
    I'm back to embarrass myself, again. The limit exists and, if I'm correct, it is zero.
     
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