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Simple Limit.

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the following Limit in terms of the number
    [tex]\alpha=\lim_{x\rightarrow 0}\frac{sinx}{x}[/tex]

    3. The attempt at a solution

    But I don't know how to convert sin(x) to sin(1/x):confused:
  2. jcsd
  3. Oct 18, 2007 #2
  4. Oct 18, 2007 #3
    I'm not terribly sure how you got


    Note that you cannot simply divide through by x when dealing with trigonometric functions. In this case you'll want to use the Squeeze Theorem.

    Note that [tex]|sin x| \leq 1[/tex]

    This implies that [tex]\left| \frac{sin x}{x} \right| \leq \frac{1}{x}[/tex]

    Now you can apply the Squeeze Theorem
  5. Oct 18, 2007 #4
    But from sqeeze theorem we get 0.
  6. Oct 18, 2007 #5


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    Gold Member

    Yeah, and what's the problem?
  7. Oct 18, 2007 #6
    Ok ,then.
  8. Oct 18, 2007 #7
    Can we in the similar manner say that [tex]\lim_{x\rightarrow\infty}\frac{cosx}{x}=0[/tex]
  9. Oct 18, 2007 #8
    Is there any point in the argument which doesn't hold as a result of switching [tex]sin(x)[/tex] to [tex]cos(x)[/tex]?
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