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Today test - one limit

  1. Jan 7, 2005 #1
    Hi all,

    could you help me with the limit we had in today test please? Here it is:

    Find out, for which [itex]\alpha \in \mathbb{R}[/itex] the limit exists and is finite:

    [tex]
    \lim_{x \rightarrow 0_{+}} \frac{ \log \left( \sqrt{1 + x^2} - x \right) - \sin \left( x^{\alpha} \right)}{x^{\alpha}}
    [/tex]

    and find the limit for those [itex]\alpha[/itex]

    I tried it, but I had too many questionable steps there, so I won't post it here till I find out what is the right solution :smile:

    Thank you.
     
  2. jcsd
  3. Jan 7, 2005 #2

    dextercioby

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    Homework Helper

    Let's take it gradually:
    Let's take the most simple case,viz.[itex] \alpha =0 [/itex]

    Can u compute the limit in that case??


    Daniel.

    PS.Provide the calculations and the result.
     
  4. Jan 7, 2005 #3
    Ok, here is how I did it:

    [tex]
    \lim_{x \rightarrow 0_{+}} \frac{ \log \left( \sqrt{1 + x^2} - x \right) - \sin \left( x^{\alpha} \right)}{x^{\alpha}} =
    [/tex]

    [tex]
    \lim_{x \rightarrow 0_{+}} \frac{ \log \left( 1 + \left(\sqrt{1 + x^2} - x - 1 \right) \right) }{x^{\alpha} \left( \sqrt{1 + x^2} - x - 1 \right)} . \lim_{x \rightarrow 0_{+}} \left( \sqrt{1 + x^2} - x - 1 \right) - \lim_{x \rightarrow 0_{+}} \frac{\sin \left( x^{\alpha} \right)}{x^{\alpha}} =
    [/tex]

    [tex]
    0.\lim_{x \rightarrow 0_{+}} \frac{1}{x^{\alpha}} - \lim_{x \rightarrow 0_{+}} \frac{\sin \left( x^{\alpha} \right)}{x^{\alpha}}
    [/tex]

    In this place I thought that [itex]\alpha[/itex] must be < 0 so that the limit will be finite (because of the [itex]\frac{1}{x^{\alpha}}[/itex] limit). Then my steps are ok and in result the limit is 0 :smile:
     
  5. Jan 7, 2005 #4

    dextercioby

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    Science Advisor
    Homework Helper

    [tex]
    L=\lim_{x \rightarrow 0_{+}} \frac{ \ln \left( \sqrt{1 + x^2} - x \right) - \sin \left( x^{\alpha} \right)}{x^{\alpha}}
    [/tex]

    Case a)[itex] \alpha =0 [/itex]
    Then
    [tex] x^{\alpha}=1;\sin(x^{\alpha})=\sin 1 [/tex]

    The limit becomes:
    [tex] L=\lim_{x\searrow 0} \ln(\sqrt{1+x^{2}}-x) -\sin 1=0-\sin 1=-\sin 1 [/tex] (1)

    Case b)[itex] \alpha >0 [/itex]
    Then u can apply August L'Ho^spital rule to get:

    [tex] \lim_{x\searrow 0} \frac{x-\sqrt{1+x^{2}}}{\alpha\sqrt{1+x^{2}}(\sqrt{1+x^{2}}-x) x^{\alpha -1}}-1 [/tex] (2)

    U have 3 cases:

    b1)[itex] \alpha>1 [/itex]
    The limit is:
    [tex] L=-\infty [/tex] (3)

    b2)[itex]\alpha=1 [/itex]
    The limit is:
    [tex] L=-1-1=-2 [/itex] (4)

    b3)[itex] 0< \alpha <1 [/itex]
    The limit is:
    [tex] L=0-1=-1 [/itex] (5)

    I let u now consider the case "c)",when [itex] \alpha < 0[/itex].

    Daniel.
     
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