Problem 47 - Calculus for Biology and Medicine

Question

In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow 0} \ln (1-x) $$

Step-by-Step Solution

Answer

0

Step 1: Identify the Limit Form

First, recognize that the expression inside the limit is \( \ln(1-x) \). We need to evaluate \( \lim_{x \to 0} \ln(1-x) \). As \( x \to 0 \), \( 1-x \to 1 \). Thus, \( \ln(1-x) \to \ln(1) = 0 \). However, we need to confirm the behavior of \( \ln(1-x) \) as \( x \to 0 \) from both sides to ensure complete understanding.

Step 2: Apply the Taylor Series Expansion

The natural logarithm function can be expanded using the Taylor series: \( \ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \ldots \). For small \( x \), we primarily consider the term \( -x \) since it dominates as \( x \to 0 \).

Step 3: Evaluate the Limit Using Dominant Term

Within the region close to zero, the logarithmic function approximates as \( \ln(1-x) \approx -x \). Consequently, evaluate the limit as: \[ \lim_{x \to 0} (-x) = 0. \] This indicates that the limit \( \lim_{x \to 0} \ln(1-x) \) indeed approaches zero from both directions.

Key Concepts

Understanding the Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a fundamental mathematical function. It is the logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. The natural logarithm has key properties that are essential in calculus and mathematical analysis.

• It is defined only for positive real numbers.
• The natural logarithm of 1 is 0, meaning \( \ln(1) = 0 \). This is because raising \( e \) to the power of 0 gives 1.
• It is a continuous and increasing function, which means as \( x \) increases, \( \ln(x) \) also increases.

The behavior of the natural logarithm ensures that operations with \( \ln(1-x) \) as \( x \) approaches 0 require careful analysis, considering it is only valid for values of \( 1-x \) that are positive.

Exploring the Taylor Series Expansion

The Taylor series expansion is a mathematical tool that represents functions as infinite sums of terms. This series can help simplify complicated functions, like the natural logarithm, into more manageable expressions for analysis.
For the function \( \ln(1-x) \), the Taylor series expansion at \( x = 0 \) is given by:\[\ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \ldots \]
This expansion is particularly useful for approximating the value of \( \ln(1-x) \) when \( x \) is very small.

• Each term in the series involves increasing powers of \( x \) and decreasing influence as \( x \) approaches 0.
• The first term, \( -x \), is the linear approximation and is dominant close to zero. This means it has the greatest impact on the value of the function when \( x \) is near 0.

The Taylor series is a powerful concept for evaluating limits because it simplifies calculations when only the dominant terms are considered.

Intuitively Approaching Zero

When examining limits, "approaching zero" refers to assessing the behavior of a function as the variable gets infinitely close to zero. For functions like \( \ln(1-x) \), understanding this concept is crucial.
In calculus, analyzing such limits helps to predict the function's behavior near specific points which might not be straightforward from the function itself.

• As \( x \to 0 \), any function of \( x \) should also approach a logical result, as seen in this case with \( \ln(1) = 0 \).
• The examination requires checking the behavior of the function from the left and right sides of zero, ensuring the function becomes appropriately small or zero from both directions as \( x \) nears zero.
• Specifically, with \( \ln(1-x) \), substituting small values of \( x \) shows that \( -x \) dominates, leading to the result of the limit \( \ln(1-x) \to 0 \).

Approaching zero and understanding limits gives us insights into the continuity and differentiability of a function, which are foundational in calculus.