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Need a hint on this limit

  • Thread starter jdstokes
  • Start date
[itex]\lim_{x\rightarrow 0^+}(1+3x)^{1/x}[/itex]

Thanks.
 
hello there

firstly i would make a substitution such as
[itex]x=\frac{1}{u}[/itex]
[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?

steven
 

HallsofIvy

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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 by a moderator:
I still don't know where to go with this. I seem to keep getting a [itex](1+ 0)^\infty[/itex] situation.
 

dextercioby

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

Daniel.
 

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