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Magazine Chapter 4: Problem 72
Use l'H么pital's Rule to evaluate the one-sided limit. \(\lim _{x \rightarrow 0^{+}} \frac{\sin (x)}{\ln (1+x)}\)
Short Answer
Expert verified
The limit is 1.
Step by step solution
01
Recognize the Indeterminate Form
Observe that as \( x \to 0^+ \), \( \sin(x) \to 0 \) and \( \ln(1+x) \to 0 \). Hence, the limit \( \lim_{x \to 0^+} \frac{\sin(x)}{\ln(1+x)} \) has the indeterminate form \( \frac{0}{0} \). This indicates that l'H么pital's Rule is applicable.
02
Apply l'H么pital's Rule
l'H么pital's Rule states that if a limit \( \lim_{x \to a} \frac{f(x)}{g(x)} \) results in the indeterminate form \( \frac{0}{0} \), the limit is the same as \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \) provided this limit exists. First, compute the derivatives: \( f(x) = \sin(x) \) with \( f'(x) = \cos(x) \) and \( g(x) = \ln(1+x) \) with \( g'(x) = \frac{1}{1+x} \).
03
Find the Limit of the Derivatives
Now evaluate the new limit: \[ \lim_{x \to 0^+} \frac{\cos(x)}{\frac{1}{1+x}} = \lim_{x \to 0^+} \cos(x) \cdot (1+x) \].
04
Evaluate the Final Expression
As \( x \to 0^+ \), \( \cos(x) \to \cos(0) = 1 \) and \( 1+x \to 1 \). Therefore, the limit of the expression becomes \( 1 \cdot 1 = 1 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
An indeterminate form occurs when the limit of a function yields an ambiguous expression, one that does not straightforwardly lead to a clear limit value. Examples include, but are not limited to, \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), and \( \infty - \infty \).
When a function approaches these forms as it approaches a limit, regular limit evaluation techniques might not work.
When a function approaches these forms as it approaches a limit, regular limit evaluation techniques might not work.
- Consider \( \lim_{x \to 0^+} \frac{\sin(x)}{\ln(1+x)} \). It results in the form \( \frac{0}{0} \) as both the numerator and denominator approach zero.
- l'H么pital's Rule is applicable here, as it is specially designed to deal with such indeterminate forms.
One-Sided Limits
A one-sided limit is the value a function approaches as the input approaches a particular value from one side only鈥攅ither from the left (-) or the right (+). One-sided limits are essential for analyzing the behavior of functions at points of discontinuity or sharp turns.
In many cases, such as with our limit problem \( \lim_{x \to 0^+} \frac{\sin(x)}{\ln(1+x)} \), we only consider inputs approaching from the positive side (right). This becomes pertinent in calculus when addressing limits at endpoints of intervals or asymptotic boundaries.
In many cases, such as with our limit problem \( \lim_{x \to 0^+} \frac{\sin(x)}{\ln(1+x)} \), we only consider inputs approaching from the positive side (right). This becomes pertinent in calculus when addressing limits at endpoints of intervals or asymptotic boundaries.
- These types of limits help refine our understanding of function behavior near boundaries and are required for asserting the existence of certain limits when continuity is not guaranteed.
- One-sided limits are often used in conjunction with l'H么pital's Rule, enhancing the ability to analyze tricky boundaries.
Derivatives
Derivatives are the foundation of calculus, representing the rate of change or slope of a function. For a function \( f(x) \), the derivative, denoted as \( f'(x) \), captures how \( f(x) \) changes for small changes in \( x \).
When applying l'H么pital's Rule, derivatives become essential, as the rule states that you can replace a limit of a quotient of functions by the limit of a quotient of their derivatives, given the initial limit is an indeterminate form \( \frac{0}{0} \).
When applying l'H么pital's Rule, derivatives become essential, as the rule states that you can replace a limit of a quotient of functions by the limit of a quotient of their derivatives, given the initial limit is an indeterminate form \( \frac{0}{0} \).
- In the exercise, the derivatives were \( f'(x) = \cos(x) \) for \( f(x) = \sin(x) \) and \( g'(x) = \frac{1}{1+x} \) for \( g(x) = \ln(1+x) \).
- Calculating these correctly is crucial to transforming and evaluating the original limit.
Calculus Limits
Limits are fundamental to the study of calculus, allowing us to analyze the behavior of functions as they approach specific points or infinity. The limit process helps define continuity, derivatives, and integrals, making them central to advanced mathematical analysis.
In the problem \( \lim_{x \to 0^+} \frac{\sin(x)}{\ln(1+x)} \), we explore the behavior of the expression as \( x \) nears zero from the positive side.
In the problem \( \lim_{x \to 0^+} \frac{\sin(x)}{\ln(1+x)} \), we explore the behavior of the expression as \( x \) nears zero from the positive side.
- The process involves examining the values functions approach rather than the values at specific points.
- Finding limits using strategies like l'H么pital's Rule is necessary when dealing with indeterminate forms.
- The computed limits allow continuity checks and derivative calculations, aiding further function analysis.
