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Need a hint on this limit

  1. Jun 15, 2005 #1
    [itex]\lim_{x\rightarrow 0^+}(1+3x)^{1/x}[/itex]

  2. jcsd
  3. Jun 15, 2005 #2
    hello there

    firstly i would make a substitution such as
    [itex]\lim_{x\rightarrow 0^+}(1+3x)^{1/x}=\lim_{u\rightarrow\infty}(1+\frac{3}{u})^{u}[/itex] you do know what this limit is equal to right?

  4. Jun 15, 2005 #3


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    It might be better to make the substitution [tex]x= \frac{1}{3u}[/tex] so that [tex]1+ 3x= 1+ \frac{1}{u}[/tex] and [tex]\frac{1}{x}= 3u[/tex]. Then the limit becomes
    [tex]\lim_{x\rightarrow 0^+}(1+3x)^{1/x}=\lim_{u\rightarrow\infty}(1+\frac{1}{u})^{3u}= \{\lim_{u\rightarrow\infty}(1+ \frac{1}{u})^u\}^3[/tex].
    Last edited: Jun 15, 2005
  5. Jun 15, 2005 #4
    I still don't know where to go with this. I seem to keep getting a [itex](1+ 0)^\infty[/itex] situation.
  6. Jun 15, 2005 #5


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    Do you know the definition of "e" (Euler's number)...?

  7. Jun 15, 2005 #6
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