We are tasked with finding the first three non-zero terms of the Maclaurin series for the function:

$f(x) = \frac{1 - e^{-x}}{x}$

Step 1: Find the Maclaurin series for $e^{-x}$

The Maclaurin series for $e^{-x}$ is:

$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots$

Step 2: Substitute the series into $f(x)$

Now, substitute the series for $e^{-x}$ into the expression for $f(x)$:

$f(x) = \frac{1 - e^{-x}}{x} = \frac{1 - \left( 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots \right)}{x}$

Simplifying the numerator:

$f(x) = \frac{1 - 1 + x - \frac{x^2}{2} + \frac{x^3}{6} - \dots}{x} = \frac{x - \frac{x^2}{2} + \frac{x^3}{6} - \dots}{x}$

Step 3: Simplify the expression

Now, divide each term in the series by $x$:

$f(x) = 1 - \frac{x}{2} + \frac{x^2}{6} - \dots$

Step 4: Conclusion

The first three non-zero terms of the Maclaurin series for $f(x) = \frac{1 - e^{-x}}{x}$ are:

$f(x) = 1 - \frac{x}{2} + \frac{x^2}{6}$

Would you like further details on how to expand this series or how to apply it? Here are some follow-up questions to explore:

1. How is the general form of a Maclaurin series derived for other functions?
2. What is the radius of convergence for the Maclaurin series of $e^{-x}$?
3. How can you use Maclaurin series to approximate functions near $x = 0$?
4. What is the difference between a Maclaurin and a Taylor series?
5. How can the error term in a truncated Maclaurin series be estimated?

Tip: The Maclaurin series is just a special case of the Taylor series, where the expansion occurs around $x = 0$.