Evaluate _x 0^+ x^ x

As $x \to 0^+$, $x \to 0$ and $\sin x \to 0$. This is an indeterminate form of type $0^0$.

Let $y = x^{\sin x}$. Then $\ln y = \ln(x^{\sin x}) = \sin x \ln x$.
Now, we evaluate $\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} \sin x \ln x$.
This is an indeterminate form of type $0 \cdot (-\infty)$.

Rewrite $\sin x \ln x$ as $\frac{\ln x}{1/\sin x} = \frac{\ln x}{\csc x}$.
Now, as $x \to 0^+$, $\ln x \to -\infty$ and $\csc x \to \infty$. This is an indeterminate form of type $\frac{-\infty}{\infty}$.

We can rewrite this as:

We know that $\lim_{x \to 0^+} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0^+} \tan x = 0$.
So, $\lim_{x \to 0^+} \ln y = -(1 \cdot 0) = 0$.

Since $\lim_{x \to 0^+} \ln y = 0$, then $\lim_{x \to 0^+} y = e^0 = 1$.

The limit is $\boxed{1}$.

2) Evaluate $\lim_{x \to 0^+} (1 + 3x)^{1/(2x)}$

As $x \to 0^+$, $1 + 3x \to 1$ and $\frac{1}{2x} \to \infty$. This is an indeterminate form of type $1^\infty$.

Recall the definition: $\lim_{u \to 0} (1 + u)^{1/u} = e$.
Let $u = 3x$. As $x \to 0^+$, $u \to 0^+$.
Substitute $u$ into the expression:

Using the property of limits for powers:

The limit is $\boxed{e^{3/2}}$.

3) Evaluate $\lim_{x \to \infty} (x^2 + x)^{1/\ln x}$

As $x \to \infty$, $x^2 + x \to \infty$ and $\ln x \to \infty$, so $\frac{1}{\ln x} \to 0$. This is an indeterminate form of type $\infty^0$.

Let $y = (x^2 + x)^{1/\ln x}$. Then $\ln y = \ln\left((x^2 + x)^{1/\ln x}\right) = \frac{1}{\ln x} \ln(x^2 + x) = \frac{\ln(x^2 + x)}{\ln x}$.
Now, we evaluate $\lim_{x \to \infty} \ln y = \lim_{x \to \infty} \frac{\ln(x^2 + x)}{\ln x}$.
As $x \to \infty$, $\ln(x^2 + x) \to \infty$ and $\ln x \to \infty$. This is an indeterminate form of type $\frac{\infty}{\infty}$.

Divide the numerator and denominator by $x^2$:

As $x \to \infty$, $\frac{1}{x} \to 0$.

So, $\lim_{x \to \infty} \ln y = 2$.

Since $\lim_{x \to \infty} \ln y = 2$, then $\lim_{x \to \infty} y = e^2$.

The limit is $\boxed{e^2}$.