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Simple Limit question

  1. Apr 12, 2005 #1
    Lim x-> infinity of
    SQRT(X)SIN(1/X)

    thank you very much
     
  2. jcsd
  3. Apr 12, 2005 #2

    dextercioby

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    Okay.

    [tex] \lim_{x\rightarrow +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{\sqrt{x}}} [/tex]

    Use the limit

    [tex] \lim_{y\searrow 0} \frac{\sin y}{y}=1 [/tex]

    and the substitution

    [tex] \frac{1}{x}=y [/tex]

    Daniel.
     
  4. Apr 12, 2005 #3
    hmm... how about a slightly easier way. We have that
    [tex]x \ge 0 \Rightarrow sin(x) \le x[/tex]
    so
    [tex]lim_{x\rightarrow \infty} \sqrt{x}\times sin(1/x) \le lim_{x\rightarrow \infty} \sqrt{x}/x=?[/tex]
    Now can you show whether or not the limit is positive? What about negative?
     
  5. Apr 12, 2005 #4
    so its = 0?
     
  6. Apr 12, 2005 #5

    dextercioby

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    That way u can show the original limit is smaller or less than 0.U need the "squeeze theorem"...,U can use

    [tex] x > 0 \Rightarrow -x <\sin x< x [/tex]

    to start with,and the afore mentioned theorem

    Daniel.
     
    Last edited: Apr 12, 2005
  7. Apr 12, 2005 #6
    I don't think I understand your complaint Daniel. All I'm saying is [tex]sin(1/x) \le 1/x[/tex] when x is positive. How does that involve the square root?
     
  8. Apr 12, 2005 #7

    dextercioby

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    Yeah,you're right.I was thinking of my post and the substitution involved there.Yes,those 2 inequalities can prove it,using "squeeze theorem".

    Daniel.
     
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