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?
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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