For the functions $f(x)=\frac{5}{x+2}$ and $g(x)=\frac{13}{x+3}$, find the composition $f \circ g$ and simplify your answer as much as possible. Write the domain using interval notation.
\[
(f \circ g)(x)=\square \quad \mid \begin{array}{ccc}
\frac{\square}{\square} & \square & \sqrt{\square} \\
\text { Domain of } f \circ g: \square & (\square, \square) & {[\square, \square]} \\
\square \cup \square & (\square, \square] & {[\square, \square)} \\
\varnothing & \infty & -\infty \\
\times & 5
\end{array}
\]

Step 1 ：Define the functions $f(x) = \frac{5}{x + 2}$ and $g(x) = \frac{13}{x + 3}$

Step 2 ：The composition of two functions, $f \circ g$, is defined as $(f \circ g)(x) = f(g(x))$. So, to find $f \circ g$, we substitute $g(x)$ into $f(x)$, which gives us $f(g(x)) = \frac{5}{2 + \frac{13}{x + 3}}$

Step 3 ：Simplify the expression to get $(f \circ g)(x) = \frac{5(x + 3)}{2x + 19}$

Step 4 ：For the domain, we consider the values of $x$ for which both $f(x)$ and $g(x)$ are defined. Both functions are defined for all real numbers except for the values that make the denominator zero. For $f(x)$, the denominator is zero when $x = -2$ and for $g(x)$, the denominator is zero when $x = -3$

Step 5 ：So, the domain of $f \circ g$ is all real numbers except $-3$ and $-2$, which can be written in interval notation as $(-\infty, -3) \cup (-3, -2) \cup (-2, \infty)$

Step 6 ：Final Answer: The composition of the functions $f$ and $g$ is $(f \circ g)(x) = \frac{5(x + 3)}{2x + 19}$ and the domain of $f \circ g$ is $(-\infty, -3) \cup (-3, -2) \cup (-2, \infty)$