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Improper - How do i do this?

  1. Oct 9, 2007 #1
    Improper -- How do i do this?

    I can't get started with the following improper limits:

    lim as x approaches infinity of (x * sin (1/x))

    I get this is in the form 0 * infinity, but when I move x to the bottom, and do l'hopital, nothing gets simpler.

    And if you have a chance,

    what is the lim as x approaches infinity of (cos x)/x ? I don't know what the limit of cos (infinity) is , so I can't get started. Thanks.
     
  2. jcsd
  3. Oct 9, 2007 #2
    cos(x) has no limit at infinity, because the function oscillates. But, it's bounded, and the denominator (x) is growing without bound. What will happen to the fraction as x gets larger?

    As for the first one, try replacing 1/x with a new variable, say w. Then, the limit as z approaches infinity of x sin (1/x) is the same as the limit as w approaches 0 of sin(w)/w, a limit you should know!
     
    Last edited: Oct 10, 2007
  4. Oct 9, 2007 #3
    -->will it approach 0 since we take cos to be 1 or a treat it as any number?

    --> for the first one, the answer is one--suing hospital or the formula sin t/t = 1.
     
  5. Oct 9, 2007 #4

    Dick

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    Yes for the sin question. For the cos, -1/x<=cos(x)/x<=1/x. Squeeze.
     
  6. Oct 10, 2007 #5

    HallsofIvy

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    You know, do you not, that [itex]\lim_{x\rightarrow 0} \frac{sin(x)}{x}= 1[/itex]?

    If so, a simple substitution should put x sin(1/x) in that form.
     
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