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

[jdstokes]

\lim_{x\rightarrow 0^+}(1+3x)^{1/x}

Thanks.

 

[steven187]

hello there

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

steven

 

[HallsofIvy]

It might be better to make the substitution x= \frac{1}{3u} so that 1+ 3x= 1+ \frac{1}{u} and \frac{1}{x}= 3u. Then the limit becomes
\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.

 

[jdstokes]

I still don't know where to go with this. I seem to keep getting a (1+ 0)^\infty situation.

 

[dextercioby]

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

Daniel.

 

[steven187]

hello there
its just a manner of recognising the limit, here is a link tells you all about the limit you should be recognising and how to prove it

http://www.meta-religion.com/Mathematics/Articles/eulers_sequence.htm

steven