Analyze the function f(x) = (e^x)/(x+2): find its domain and limits at the boundaries of the domain.

Here's a detailed analysis of the function $f(x) = \frac{e^x}{x+2}$.

a) The domain of $f(x)$
The function $f(x)$ is defined for all real numbers where the denominator is not zero.
The denominator is $x+2$.
Setting the denominator to zero:

$$x+2 = 0$$ $$x = -2$$

Thus, the function is defined for all real numbers except $x=-2$.
The domain is $\boxed{\mathbb{R} \setminus \{-2\}}$.

b) The limits at the boundaries of the domain
The boundaries of the domain are $x \to -\infty$, $x \to -2^-$, $x \to -2^+$, and $x \to \infty$.

1. As $x \to -\infty$:

   $$\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{e^x}{x+2}$$

   As $x \to -\infty$, $e^x \to 0$ and $x+2 \to -\infty$.

   $$\lim_{x \to -\infty} f(x) = \frac{0}{-\infty} = 0$$

2. As $x \to -2^-$ (from the left side of $-2$):

   $$\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} \frac{e^x}{x+2}$$

   As $x \to -2^-$, $e^x \to e^{-2}$ (a positive constant) and $x+2 \to 0^-$ (a small negative number).

   $$\lim_{x \to -2^-} f(x) = \frac{e^{-2}}{0^-} = -\infty$$

3. As $x \to -2^+$ (from the right side of