Problem 47
Question
In Problems 29-48, find the limits. $$ \lim _{x \rightarrow 0} \ln (1-x) $$
Step-by-Step Solution
Verified Answer
The limit is 0.
1Step 1: Recognize the Indeterminate Form
As \( x \) approaches 0, \( 1-x \) approaches 1, and \( \ln(1-x) \) approaches \( \ln(1) \), which equals 0. The direct substitution gives \( \ln(1-0) = \ln(1) = 0 \). This is not an indeterminate form; hence, further analysis is unnecessary. The limit can be directly evaluated.
2Step 2: Direct Substitution
Since the function is \( \ln(1-x) \), substitute \( x = 0 \) directly into the function. We get:\[\lim_{{x \to 0}} \ln(1-x) = \ln(1-0) = \ln(1) = 0\]This confirms the value obtained from recognizing that \( \ln(1) = 0 \) in the previous step.
Key Concepts
Indeterminate FormsDirect SubstitutionNatural Logarithm
Indeterminate Forms
In calculus, an indeterminate form refers to a situation where you cannot determine the limit of a function just by simple substitution. These forms often arise when evaluating limits, and they require more analysis beyond direct substitution. Some common examples of indeterminate forms are \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( \infty - \infty \), and more.
However, not every limit problem results in an indeterminate form. In the provided exercise, evaluating \( \lim _{x \rightarrow 0} \ln (1-x) \) does not lead to an indeterminate form. When \( x \) approaches 0, \( 1-x \) approaches 1, and thus \( \ln(1-x) \) approaches \( \ln(1) \), which equals 0. Since no problematic forms like \( \frac{0}{0} \) or \( \infty - \infty \) occur, you can directly evaluate the limit without complex manipulations.
However, not every limit problem results in an indeterminate form. In the provided exercise, evaluating \( \lim _{x \rightarrow 0} \ln (1-x) \) does not lead to an indeterminate form. When \( x \) approaches 0, \( 1-x \) approaches 1, and thus \( \ln(1-x) \) approaches \( \ln(1) \), which equals 0. Since no problematic forms like \( \frac{0}{0} \) or \( \infty - \infty \) occur, you can directly evaluate the limit without complex manipulations.
Direct Substitution
Direct substitution is a straightforward method of evaluating limits by simply substituting the given \( x \) value into the function. This method is typically used when substituting \( x \) does not result in an indeterminate form. In the given exercise, direct substitution is useful because the function is well-behaved as \( x \) approaches 0.
By substituting \( x = 0 \) directly into \( \ln(1-x) \), we get \( \ln(1-0) = \ln(1) \). Since the natural logarithm of 1 is 0, the limit directly evaluates to 0 without any further steps needed. This simplicity can be attributed to the absence of any indeterminate form in this context.
Direct substitution often provides a quick and efficient means of finding limits, especially when the function maintains its form and doesn’t lead to ambiguity at the point of interest.
By substituting \( x = 0 \) directly into \( \ln(1-x) \), we get \( \ln(1-0) = \ln(1) \). Since the natural logarithm of 1 is 0, the limit directly evaluates to 0 without any further steps needed. This simplicity can be attributed to the absence of any indeterminate form in this context.
Direct substitution often provides a quick and efficient means of finding limits, especially when the function maintains its form and doesn’t lead to ambiguity at the point of interest.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It is a fundamental function in calculus, particularly useful in simplifying the integration and differentiation of exponential functions.
In the context of limits, understanding the behavior of the natural logarithm near specific points is essential. For example, \( \ln(1) = 0 \) is a crucial property, as demonstrated in the exercise where substituting \( \ln(1-x) \) into the limit reveals that the function approaches \( \ln(1) \) as \( x \) approaches 0.
This property simplifies solving limits involving natural logarithms because it eliminates the need for complex limit laws or L'Hôpital's Rule, especially when the function does not descend into an indeterminate form. Being familiar with the basic properties of the natural logarithm can significantly ease the process of solving calculus problems.
In the context of limits, understanding the behavior of the natural logarithm near specific points is essential. For example, \( \ln(1) = 0 \) is a crucial property, as demonstrated in the exercise where substituting \( \ln(1-x) \) into the limit reveals that the function approaches \( \ln(1) \) as \( x \) approaches 0.
This property simplifies solving limits involving natural logarithms because it eliminates the need for complex limit laws or L'Hôpital's Rule, especially when the function does not descend into an indeterminate form. Being familiar with the basic properties of the natural logarithm can significantly ease the process of solving calculus problems.
Other exercises in this chapter
Problem 46
In Problems 29-48, find the limits. $$ \lim _{x \rightarrow 0} \frac{5-\sqrt{25+x^{2}}}{2 x^{2}} $$
View solution Problem 46
In Problems 39-56, use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow 1} \frac{x^{3}+1}{x+2} $$
View solution Problem 47
In Problems 39-56, use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow 3} \frac{3 x^{2}+1}{2 x-3} $$
View solution Problem 48
In Problems 29-48, find the limits. $$ \lim _{x \rightarrow 1} \ln \left[e^{x} \cos (x-1)\right] $$
View solution