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Limit proof

  1. Mar 28, 2005 #1
    If we are asked to prove the following: [tex] \lim_{x\rightarrow 0} \frac{x^{2}\sin\frac{1}{x}}{\sin x} = 0 [/tex] would I do: [tex] \lim_{x\rightarrow 0} x^{2}(\sin\frac{1}{x})(\frac{1}{\sin x}) [/tex]? So the leading factor (dampening factor) approaches 0 which makes the whole expression approach 0?

    Thanks
     
  2. jcsd
  3. Mar 28, 2005 #2

    robphy

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    It's probably better to use L'Hopital's Rule.

    Note that you see some expressions of the form (sin y)/y, which may help you.
     
  4. Mar 28, 2005 #3

    dextercioby

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    Use the limits

    [tex] \lim_{x\rightarrow 0}\frac{\sin x}{x}=1 [/tex]

    [tex] \lim_{x\rightarrow +\infty} \frac{\sin x}{x}=0 [/tex]

    Daniel.
     
  5. Mar 28, 2005 #4
    So you have: [tex] \lim_{x\rightarrow 0} x^{2}(\frac{\sin x}{x})(\frac{1}{\sin x}) [/tex] . And you get [tex] \frac{x^{2}}{\sin x} [/tex]. So the limit is then 0?
     
    Last edited: Mar 28, 2005
  6. Mar 28, 2005 #5

    robphy

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    Check your algebra:

    [tex] \lim_{x\rightarrow 0} \frac{x^{2}\sin(\frac{1}{x})}{\sin x} =
    \lim_{x\rightarrow 0} \frac{x}{\sin x} \frac{\sin(\frac{1}{x})}{\frac{1}{x}}
    [/tex]
     
  7. Mar 29, 2005 #6

    dextercioby

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    The limit is 0,but the "sines" cannot be simplified.It's a product of 2 fractions,one o them goes to 1 & the other to 0.

    Daniel.
     
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