Chapter 1: Problem 41
Challenge your understanding of limits but can be evaluated using the knowledge gained in this section. $$ \lim _{x \rightarrow 0} \frac{\ln (1+x)}{x} $$
Short Answer
Expert verified
The limit is 1.
Step by step solution
01
Identify the Indeterminate Form
First, substitute \( x = 0 \) into the expression \( \frac{\ln(1+x)}{x} \) to see if it forms an indeterminate. We get \( \frac{\ln(1+0)}{0} = \frac{\ln(1)}{0} = \frac{0}{0} \), which is an indeterminate form indicating the use of L'Hôpital's Rule or other methods is necessary.
02
Apply L'Hôpital's Rule
Since the limit results in an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) if the limit is indeterminate. Here, \( f(x) = \ln(1+x) \) and \( g(x) = x \).
03
Differentiate the Numerator and Denominator
Find the derivative of \( f(x) = \ln(1+x) \), which is \( f'(x) = \frac{1}{1+x} \). Find the derivative of \( g(x) = x \), which is \( g'(x) = 1 \).
04
Evaluate the Limit of the Derivatives
Substitute the derivatives into L'Hôpital's Rule: \( \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{1}{1+x} \). Evaluate this limit by plugging \( x = 0 \): \( \frac{1}{1+0} = 1 \).
05
Conclusion
Therefore, the original limit \( \lim_{x \to 0} \frac{\ln(1+x)}{x} \) evaluates to \( 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
In calculus, we often encounter expressions whose limits cannot be determined just by substitution. These expressions are known as "indeterminate forms." One common indeterminate form is \( \frac{0}{0} \). Imagine having an equation where both the numerator and the denominator evaluate to zero when a particular value of \( x \) is substituted. Since dividing zero by zero does not produce a clear result, this form is called indeterminate.
When you see \( \frac{0}{0} \), it suggests that a direct calculation can lead to multiple possible outcomes. In such cases, calculus techniques like L'Hôpital's Rule come to our rescue to solve the limit and find a definite answer.
In our example, we could not directly compute the limit of \( \frac{\ln(1+x)}{x} \) as \( x \to 0 \) because both \( \ln(1+x) \) and \( x \) approach zero, forming \( \frac{0}{0} \). This is why identifying indeterminate forms is a crucial first step when evaluating limits in calculus.
When you see \( \frac{0}{0} \), it suggests that a direct calculation can lead to multiple possible outcomes. In such cases, calculus techniques like L'Hôpital's Rule come to our rescue to solve the limit and find a definite answer.
In our example, we could not directly compute the limit of \( \frac{\ln(1+x)}{x} \) as \( x \to 0 \) because both \( \ln(1+x) \) and \( x \) approach zero, forming \( \frac{0}{0} \). This is why identifying indeterminate forms is a crucial first step when evaluating limits in calculus.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful method used to solve indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter such forms, L'Hôpital's Rule allows you to take the derivative of the numerator and the derivative of the denominator to find the limit. Essentially, it simplifies the problem by transforming it into something more manageable.
Here's how it works: If you have a limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) that results in an indeterminate form, you can apply L'Hôpital's Rule provided both \( f(x) \) and \( g(x) \) are differentiable. The rule states that you can calculate \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) instead, where \( f'(x) \) and \( g'(x) \) are the derivatives of \( f(x) \) and \( g(x) \) respectively.
For our given problem, we used L'Hôpital's Rule on the limit \( \lim_{x \to 0} \frac{\ln(1+x)}{x} \). After differentiating, we had the simpler limit \( \lim_{x \to 0} \frac{1}{1+x} \), which evaluated cleanly to 1.
Here's how it works: If you have a limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) that results in an indeterminate form, you can apply L'Hôpital's Rule provided both \( f(x) \) and \( g(x) \) are differentiable. The rule states that you can calculate \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \) instead, where \( f'(x) \) and \( g'(x) \) are the derivatives of \( f(x) \) and \( g(x) \) respectively.
For our given problem, we used L'Hôpital's Rule on the limit \( \lim_{x \to 0} \frac{\ln(1+x)}{x} \). After differentiating, we had the simpler limit \( \lim_{x \to 0} \frac{1}{1+x} \), which evaluated cleanly to 1.
Natural Logarithm
The natural logarithm, represented as \( \ln(x) \), is a crucial mathematical function often encountered in calculus and higher mathematics. Its base is the irrational number \( e \approx 2.71828 \), and it provides a way to solve exponential growth problems and logarithmic equations.
Let's look at some properties of the natural logarithm:
The natural logarithm also plays a vital role in various fields like statistics, physics, and finance, where it models naturally occurring phenomena, growth rates, and decay processes.
Let's look at some properties of the natural logarithm:
- \( \ln(1) = 0 \) because \( e^0 = 1 \).
- It is defined only for positive values of \( x \).
- It grows unbounded as \( x \) increases and decreases towards negative infinity as \( x \) approaches zero from the positive side.
The natural logarithm also plays a vital role in various fields like statistics, physics, and finance, where it models naturally occurring phenomena, growth rates, and decay processes.