Evaluate the limit: _x 0^+ (1 + 3x)^(1)/(2x)

Let $u = 3x$. As $x \to 0^+$, $u \to 0^+$.
We need the exponent to be in terms of $u$. The original exponent is $\frac{1}{2x}$.
Since $u = 3x$, we have $x = \frac{u}{3}$.
Substitute $x = \frac{u}{3}$ into the exponent:

$$\frac{1}{2x} = \frac{1}{2\left(\frac{u}{3}\right)} = \frac{1}{\frac{2u}{3}} = \frac{3}{2u}$$

We can rewrite the exponent $\frac{3}{2u}$ as $\frac{1}{u} \cdot \frac{3}{2}$.

$$\lim_{u \to 0^+} (1 + u)^{\frac{1}{u} \cdot \frac{3}{2}}$$

Using the exponent rule $(a^b)^c = a^{bc}$, we can write this as:

$$\lim_{u \to 0^+} \left( (1 + u)^{\frac{1}{u}} \right)^{\frac{3}{2}}$$

Since $\lim_{u \to 0^+} (1 + u)^{\frac{1}{u}} = e$, we substitute this into the expression:

$$e^{\frac{3}{2}}$$