Use Maclaurin series to evaluate the limits. \(\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)\) Hint: Combine the fractions.

Limits are essential in calculus. They help us understand the behavior of functions as inputs approach specific points. In this exercise, we evaluate the limit as x approaches 0:

o(\(\frac{1}{x} - \frac{1}{e^x - 1}\)).
Understanding this expression involves calculating how these two terms approach their values as x gets very close to zero. Always pay attention to small increments. They often simplify complex function behavior.
In calculus, limits provide a foundation to discover derivatives and integrals by revealing function behaviors at critical points.

The exponential function, denoted as \(e^x\), is crucial in mathematical analysis. It grows rapidly and appears frequently in various scientific contexts. Its key property is that its derivative is the same as the function itself.
In the Maclaurin series, the exponential function is expanded as:
e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \text{higher-order terms}

In our context, we use this series because it approximates the function around x=0. The series is infinitely long, but only a few terms often provide a good approximation for small values of x. Understanding this series helps simplify and solve complex limits involving \(e^x\).

The Taylor series generalizes the Maclaurin series, giving us a way to represent functions as infinite sums of their derivatives' values at a specific point. When this point is zero, it is called the Maclaurin series.
For the exponential function \(e^x\), we already saw the series:
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \text{higher terms}

This expansion allows us to represent complex functions more simply. Using the first few terms often suffices to approximate the function very closely, especially when x is small. This approach simplifies calculations and reveals underlying patterns in functions.

Combining fractions is a powerful algebraic tool to simplify expressions. In this exercise, we have the fraction:
o(\(\frac{1}{x} - \frac{1}{e^x - 1}\))
Combining them into a single fraction simplifies our calculations. Here’s how we do it:
o(\(\frac{1}{x} - \frac{1}{e^x - 1} = \frac{e^x - 1 - x}{x(e^x - 1)}\))

By substituting the Maclaurin series for \(e^x - 1\), we get:
e^x - 1 = x + \frac{x^2}{2!} + \text{higher-order terms}

After simplifying, we see that the higher-order terms become negligible as x approaches 0:
o(\(\frac{\frac{x^2}{2} + \text{higher terms}}{x^2 + \text{higher terms}} \rightarrow \frac{1}{2}\))

Hence, the complex limit reduces easily to 1/2 using fraction combinations and Maclaurin series expansion.