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Simple Limit

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data

    ok, this is the limit i want to calculate: limit x->0 sin(1/x) / (1/x) and

    limit x->infinity sin(1/x) / (1/x)

    can you tell me the clear step mathematical proof



    2. The attempt at a solution

    now the first one i tired doing this, let u =(1/x) then: limit U->infinity sin(U) / (U)

    now i understand its actually zero since, sin(U) goes back between -1 and 1, when u grows to infinity so its zero...so what's the mathematical proof.
     
    Last edited: Apr 16, 2010
  2. jcsd
  3. Apr 15, 2010 #2

    Char. Limit

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    Gold Member

    You can't take the limit as x goes to infinity of sin(u)/u...

    You need to find something else.

    As x goes to infinity, what does u go to?

    Because that's the limit you need.
     
  4. Apr 16, 2010 #3
    My mistake it was suppose to be u, since u=1/x...as x->0....u->infinity
     
  5. Apr 16, 2010 #4

    Char. Limit

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    But in the problem you were working on last, x wasn't going to zero. x was going to infinity.

    However, yes, that will be useful in the second problem.
     
  6. Apr 16, 2010 #5
    I can tell for the first one you're nearly there... after you've rewritten lim (x to 0) sin(1/x) / (1 / x) with u = 1/x, we have lim (u to infinity) sin (u) / u.

    In this case, how does the Squeeze theorem apply?

    For the second case, you might consider L' Hospital's rule, or the power series expansion of sine.
     
  7. Apr 16, 2010 #6
    for the second case:

    limit x->infinity sin(1/x) / (1/x) if i appy l'hospital...

    limit x->infinity (-1/x^2) (cos(1/x)) / (-1/x^2)

    thus i get limit x->infinity cos(1/x) and again stuck.
     
  8. Apr 16, 2010 #7
    Why are you stuck? If x goes to infinity, then 1/x goes to 0, and cos(0) is well defined :)
     
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