Question

Use Maclaurin series to evaluate the limits. \(\lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{x^{2}}-\frac{1}{x}\right)\)

Short answer

-\frac{1}{2}

Step-by-step solution

Recall the Maclaurin series for \( \ln(1+x) \)

The Maclaurin series for \( \ln(1+x) \) is \( \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\text{higher order terms} \).

Substitute the series into the limit expression

Substitute \( \ln(1+x) \) from the Maclaurin series: \[ \lim_{x \rightarrow 0}\bigg( \frac{\ln(1+x)}{x^2} - \frac{1}{x} \bigg) = \lim_{x \rightarrow 0}\bigg( \frac{x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots}{x^2} - \frac{1}{x} \bigg). \]

Simplify the expression in the limit

Split the series inside the limit into individual fractions: \[ \lim_{x \rightarrow 0}\bigg( \frac{x}{x^2} - \frac{\frac{x^2}{2}}{x^2} + \frac{\frac{x^3}{3}}{x^2} - \cdots - \frac{1}{x} \bigg) = \lim_{x \rightarrow 0}\bigg( \frac{1}{x} - \frac{1}{2} + \frac{x}{3} - \cdots - \frac{1}{x} \bigg). \]

Cancel terms and evaluate the limit

Observe that \( \frac{1}{x} \) and \( -\frac{1}{x} \) cancel out: \[ \lim_{x \rightarrow 0}\bigg( \frac{1}{x} - \frac{1}{x} - \frac{1}{2} + \frac{x}{3} - \cdots \bigg) = \lim_{x \rightarrow 0}\bigg( -\frac{1}{2} + \frac{x}{3} - \cdots \bigg) = -\frac{1}{2}. \]

Key concepts

Maclaurin Series

The Maclaurin series is a specific kind of Taylor series that expands a function around the point zero. It allows us to express many functions as an infinite sum of terms. Each term involves the function's derivatives evaluated at zero.
For example, the Maclaurin series for \(\ln(1+x)\) is:\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \.\]
This expression helps us to compute limits and approximate values for small values of \(x\).
By using the Maclaurin series, we can turn complicated functions into simpler polynomial forms, making it easier to solve problems analytically.

Evaluating Limits

Limits analyze what happens to a function as its input approaches a particular value. This concept is fundamental in calculus, often used to find the behavior of functions at specific points or as input values get very large or small.
Here, we use the limit:\[ \lim_{x \to 0}\left(\frac{\ln(1+x)}{x^2} - \frac{1}{x}\right) \.\]
By substituting the Maclaurin series of \(\ln(1+x)\), we can express the function in a simpler form:\[ \ln(1+x) \approx x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \.\]
In the process, higher-order terms become negligible as \(x\) approaches zero. We derive the limit by cancelling out similar terms and simplifying what's left.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is approximately 2.71828. It is a crucial function in mathematics with applications in calculus, economics, and sciences.
The function \(\ln(1+x)\) particularly is useful for approximations and limits, as it can be expanded using the Maclaurin series. This makes complex limits more approachable.
In our problem, we utilized the Maclaurin series for \(\ln(1+x)\) to evaluate:\[ \lim_{x \to 0}\left(\frac{\ln(1+x)}{x^2} - \frac{1}{x}\right)\]
so that we can transform it into a polynomial form and easily find the limit value.