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Magazine Chapter 4: Problem 71
Use l'H么pital's Rule to evaluate the one-sided limit. \(\lim _{x \rightarrow 0^{+}} \frac{\ln (1+x)}{\ln (1+3 x)}\)
Short Answer
Expert verified
The one-sided limit evaluates to \(\frac{1}{3}\).
Step by step solution
01
Identifying the Indeterminate Form
First, we need to evaluate the limit \( \lim_{x \to 0^+} \frac{\ln(1+x)}{\ln(1+3x)} \) to see if it results in an indeterminate form. As \(x\) approaches 0 from the positive side, both \(\ln(1+x)\) and \(\ln(1+3x)\) approach \(\ln(1) = 0\). Therefore, the expression yields the indeterminate form \(\frac{0}{0}\). This allows us to apply l'H么pital's Rule.
02
Apply l'H么pital's Rule
l'H么pital's Rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \text{ or } \frac{\pm\infty}{\pm\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), given the new limit exists. We differentiate the numerator and denominator:- The derivative of \(\ln(1+x)\) is \( \frac{1}{1+x} \).- The derivative of \(\ln(1+3x)\) is \( \frac{3}{1+3x} \).Thus, the limit becomes:\[\lim_{x \to 0^+} \frac{\frac{1}{1+x}}{\frac{3}{1+3x}} = \lim_{x \to 0^+} \frac{1+3x}{3(1+x)}\]
03
Simplify and Evaluate the New Limit
Simplify the expression from the previous step:\[\lim_{x \to 0^+} \frac{1+3x}{3(1+x)} = \lim_{x \to 0^+} \frac{1+3x}{3+3x}.\]Substitute \(x = 0\) into the expression:\[\frac{1+3(0)}{3+3(0)} = \frac{1}{3}.\]The limit evaluates to \(\frac{1}{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Evaluation
Evaluating limits is a fundamental concept in calculus. It involves finding the value that a function approaches as the input approaches a certain point. In this exercise, we're looking at the limit \( \lim_{x \rightarrow 0^{+}} \frac{\ln(1+x)}{\ln(1+3x)} \), which presents a one-sided limit. The notation \( x \to 0^+ \) means we're considering values of \( x \) that approach zero from the positive side. Evaluating this type of limit often requires special techniques, especially when indeterminate forms appear.
- First, determine if substitution gives a clear result. If not, advanced methods like L'H么pital's Rule might be necessary.
- In one-sided limits, pay attention to how the behavior of the function changes when approaching from the chosen side.
Indeterminate Forms
Indeterminate forms occur when direct substitution in a limit evaluation doesn't provide a clear solution. Commonly, these forms include \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).In our exercise, substituting \( x = 0 \) into \( \frac{\ln(1+x)}{\ln(1+3x)} \) gives \( \frac{0}{0} \), signaling an indeterminate form. This is a typical scenario where L'H么pital's Rule becomes useful, allowing us to differentiate both the numerator and the denominator.
- Indeterminate forms require careful handling and often need specific techniques like L'H么pital's Rule for resolution.
- These forms indicate that the function's behavior isn't straightforward, requiring further manipulation to evaluate the limit.
Derivatives
Derivatives are a core concept in calculus, representing the rate of change of a function. They are crucial for evaluating indeterminate forms via L'H么pital's Rule.In this context:
- The derivative of \( \ln(1+x) \) is \( \frac{1}{1+x} \). This derivative tells us how fast the function \( \ln(1+x) \) changes as \( x \) changes.
- Similarly, the derivative of \( \ln(1+3x) \) is \( \frac{3}{1+3x} \). This gives us the rate of change of the function \( \ln(1+3x) \).
Calculus
Calculus, the mathematical study of change, lies at the heart of analyzing and understanding complex functions and limits. This branch of mathematics is divided into two major parts: differential calculus and integral calculus.
- Differential calculus focuses on derivatives, which tell us about the rate of change and are essential for solving limit problems involving indeterminate forms.
- Integral calculus is concerned with accumulation, which is another critical aspect of analyzing functions.
