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Limits w/ l'hospital

  1. Jan 27, 2007 #1
    1. The problem statement, all variables and given/known data
    Compute the limit [tex]\lim_{x \rightarrow 0} (1 + 3x)^{3/x}[/tex].

    2. Relevant equations
    3. The attempt at a solution

    [tex]\lim_{x \rightarrow 0} (1 + 3x)^{3/x} = \lim_{x \rightarrow 0} e^{\frac{3}{x} \ln (1 + 3x)}[/tex]
    [tex]\lim_{x \rightarrow 0} \frac{3}{x} \ln (1 + 3x) = 3 \lim_{x \rightarrow 0} \frac{ln (1 + 3x)}{x}[/tex]
    [tex]= 3 \lim_{x \rightarrow 0} \frac{\frac{3}{1 + 3x}}{1}[/tex]
    [tex]= 3 \lim_{x \rightarrow 0} \frac{3}{1 + 3x}[/tex]
    [tex]=9[/tex]
    [tex]\lim_{x \rightarrow 0} (1 + 3x)^{3/x} = e^9[/tex]

    I just need to know if this is right.
     
    Last edited: Jan 27, 2007
  2. jcsd
  3. Jan 27, 2007 #2

    Dick

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    Homework Helper

    It is just fine.
     
  4. Jan 27, 2007 #3
    The answer is right. Just plug in .01 or .001 into the limit using the calculator and compare it to the exact value.
     
  5. Jan 27, 2007 #4

    Gib Z

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    Homework Helper

    Or You could have manipulated the definition of the exponential: [tex]e^a=\lim_{x \rightarrow \inf} (1+\frac{a}{x})^x[/tex]

    Rewrite [tex]\lim_{x \rightarrow 0} (1 + 3x)^{3/x}[/tex] as [tex]\lim_{x \rightarrow 0} (1 + \frac{3}{1/x})^{3(1/x)}[/tex]
    The limit of 1/x as 0 approaches zero is the same as x appracohes infinity. So we can rewrite the limit as
    [tex]\lim_{x \rightarrow \inf} (1 + 3/x)^{3x} = \lim_{x \rightarrow \inf}( (1 + 3/x)^x )^3[/tex] which is the definiton of e^3, then cubed, getting out e^9 as required.
     
    Last edited: Jan 27, 2007
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