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Simple limit

  1. Nov 27, 2007 #1
    Could someone please, step by step, compute the limit as x goes to 0, x >0, of:


    I am confused about it. Thanks.
  2. jcsd
  3. Nov 27, 2007 #2


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  4. Nov 27, 2007 #3
    If you know 1/x goes to [tex]\infty[/tex] then you can use a comparison test [tex]\frac{1}{\sqrt{x}} \geq \frac{1}{x}[/tex] for 0 < x < 1.

    Otherwise you have to apply the definition of a limit tending to infinity..

    For [tex]0 < x < \frac{1}{M^2}[/tex] you have [tex]\frac{1}{\sqrt{x}} > M[/tex].
  5. Nov 28, 2007 #4


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    It's not clear what you mean by "compute" the limit. Since x is in the denominator,I would think it obvious that, as x goes to 0, the fraction goes to infinity.
  6. Nov 28, 2007 #5


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    Then [tex]\sqrt{x} \leq x[/tex] for 0<x<1 which means [tex]x \leq x^2[/tex] which gives [tex]1 \leq x[/tex]
  7. Nov 28, 2007 #6
    Correction noted: 1/sqrt(x) <= 1/x. Comparison test doesn't work.
  8. Nov 28, 2007 #7
    [tex]\lim \limits_{x \to 0^ + } \frac{1}{{2\sqrt x }} = 0\because \forall \varepsilon > 0,\;\exists \delta > 0:0 < x < \delta \to \frac{1}
    {{2\sqrt x }} < \varepsilon ,\;{\text{simply choose }}\delta = \frac{1}
    {{4\varepsilon ^2 }}[/tex]
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