Question

Use Maclaurin series to evaluate the limits. \(\lim _{x \rightarrow 0} \frac{\ln (1-x)}{x}\)

Short answer

-1

Step-by-step solution

Write the Maclaurin series for \(\text{ln}(1-x)\)

The Maclaurin series for \(\text{ln}(1-x)\) is given by \(\text{ln}(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \frac{x^5}{5} - \frac{x^6}{6} - \frac{x^7}{7} - ...\)

Substitute the series into the limit expression

We need to evaluate the given limit \(\frac{\text{ln}(1-x)}{x}\). Substitute the Maclaurin series into this expression: \(\frac{-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - ...}{x}\).

Simplify the expression

Divide each term in the numerator by \(x\): \(\frac{\text{ln}(1-x)}{x} = \frac{-x}{x} - \frac{\frac{x^2}{2}}{x} - \frac{\frac{x^3}{3}}{x} - ...\) simplifies to \(-1 - \frac{x}{2} - \frac{x^2}{3} - \frac{x^3}{4} - ...\).

Evaluate the limit as \(x \rightarrow 0\)

As \(x \rightarrow 0\), all the terms involving \(x\) in the expression \(-1 - \frac{x}{2} - \frac{x^2}{3} - ...\) become 0. Therefore, the limit is \(-1\).

Key concepts

limits in calculus

Limits in calculus help us understand the behavior of functions as they approach a certain point. They are essential for defining derivatives and integrals. When we say \(\text{x} \rightarrow a\), it means \(\text{x}\) is getting very close to \(a\). We use limits to evaluate expressions like \(\frac{\text{ln}(1-x)}{x}\) as \(x\) approaches 0. To find this limit, you replace the function with a series expansion or another known form to make it simpler to evaluate. Limits help in determining the values that a function tends to, making them a core part of calculus.

natural logarithm

The natural logarithm, denoted by \( \text{ln}(x) \), is the logarithm to the base \(\text{e} \), where \(\text{e}\) is an irrational constant approximately equal to 2.71828. For small values of \:\(x\), there is a series expansion for \(\text{ln}(1-x)\) that we can use. The Maclaurin series for \(\text{ln}(1-x)\) is \(\text{ln}(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - ... \). This series provides a way to approximate \(\text{ln}(1-x)\) for small values of \(\text{x}\), which simplifies the evaluation of limits involving \(\text{ln}(1-x)\).

series expansion

A series expansion represents a function as an infinite sum of terms. The Maclaurin series is a type of series expansion that approximates functions near \(x = 0\). For example, the Maclaurin series for \(\text{ln}(1-x) \) is \(\text{ln}(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - ... \). To use it for limits, you substitute the series into the function to be evaluated. This turns complex expressions into simpler ones that are easier to handle. Series expansions are powerful tools in calculus and approximate functions when their exact forms are difficult to work with.

calculus

Calculus is a branch of mathematics focused on change and motion. It includes derivative and integral rules, helping in understanding slopes of curves and areas under them. In calculus, we often deal with limits, derivatives, and integrals. For example, evaluating a limit involves understanding how a function behaves as it approaches a specific point. We use this in the given problem to understand the behavior of \(\frac{\text{ln}(1-x)}{x} \) as \( x \rightarrow 0 \). Here, calculus principles allow us to break down and simplify such problems using methods like series expansions.