What is the Limit of (1+3x)^(1/x) as x Approaches 0+?

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Homework Help Overview

The discussion revolves around evaluating the limit of the expression (1+3x)^(1/x) as x approaches 0 from the positive side. This involves concepts from calculus, particularly limits and exponential functions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants suggest different substitutions to transform the limit into a more manageable form, such as x = 1/u and x = 1/(3u). There is also a mention of encountering an indeterminate form (1+0)^\infty, which raises questions about how to proceed from that point.

Discussion Status

Some participants have provided substitutions that may simplify the limit, while others express uncertainty about how to resolve the resulting forms. There is an ongoing exploration of the definitions and properties of limits, particularly in relation to Euler's number.

Contextual Notes

Participants are navigating through the complexities of limits and exponential functions, with some referencing external resources for further clarification. The discussion reflects a mix of understanding and confusion regarding the appropriate approaches to take.

jdstokes
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[itex]\lim_{x\rightarrow 0^+}(1+3x)^{1/x}[/itex]

Thanks.
 
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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
 
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].
 
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I still don't know where to go with this. I seem to keep getting a [itex](1+ 0)^\infty[/itex] situation.
 
Do you know the definition of "e" (Euler's number)...?

Daniel.
 

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