The following functions have no Maclaurin expansions since they are not defined at x = 0 except;

To determine which functions have Maclaurin expansions, we need to recall that a Maclaurin series is a Taylor series centered at $x = 0$. A function must be defined and have continuous derivatives at that point to have a Maclaurin series. Here’s how we can approach this problem:

1. Break Down the Problem

   • Identify functions that are not defined at $x = 0$.
   • Determine if any of the given functions can actually be evaluated and/or have derivatives at $x = 0$.

2. Relevant Concepts

   • A Maclaurin series for a function $f(x)$ is given by: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$
   • A function must be continuous at $x = 0$ and have defined derivatives at that point for the series to exist.

3. Analysis and Detail

   • Examine the functions in question to see if they are defined at $x = 0$. Common candidates that might be undefined include functions like $\frac{1}{x}$ or $\ln(x)$, which are not defined at $x = 0$.
   • Consider functions whose limits can yield values at $x = 0$, for example, $\lim_{x \to 0} e^{-x^2} = 1$, meaning it has a defined value and derivatives at that point.

4. Verify and Summarize

   • Go through each function to either confirm or deny that they have Maclaurin series based on their definitions and behavior at $x = 0$.
   • Summarize that functions which are defined at $x = 0$ can have Maclaurin expansions, while those undefined cannot.

Final Answer

The function that has a Maclaurin expansion is the one that is defined at $x = 0$. Common examples like $e^{x}$, $\sin(x)$, and $\cos(x)$ have Maclaurin expansions, while functions like $\frac{1}{x}$ and $\ln(x)$ do not. You will determine the specific function from your list that aligns with this criterion.