Chapter 27: Problem 32
Evaluate each limit (if it exists). Use \(L\) Hospital's rule (if appropriate). $$\lim _{x \rightarrow 0^{+}}\left(\sin ^{-1} x\right)(\ln x)$$
Short Answer
Expert verified
The limit is -1.
Step by step solution
01
Identify the Limit Form
We need to evaluate the limit \( \lim_{x \to 0^+} (\sin^{-1} x)(\ln x) \). First, observe the form of this expression as \( x \to 0^+ \). \( \sin^{-1} x \approx x \) for small \( x \), and \( \ln x \to -\infty \). Thus, the expression is of the indeterminate form \( 0 \times -\infty \).
02
Rewrite the Expression
To apply L'Hôpital's Rule, we need to rewrite the expression into a fraction. We rewrite the product \( (\sin^{-1} x)(\ln x) \) as a fraction: \( \frac{\ln x}{\frac{1}{\sin^{-1} x}} \). As \( x \to 0^+ \), \( \ln x \to -\infty \) and \( \frac{1}{\sin^{-1} x} \to +\infty \), so we now have the form \( \frac{-\infty}{\infty} \), which is appropriate for L'Hôpital's Rule.
03
Apply L'Hôpital's Rule
According to L'Hôpital's Rule, if a limit is in the form \( \frac{\infty}{\infty} \), we can differentiate the numerator and the denominator separately. We differentiate \( \ln x \) to get \( \frac{1}{x} \) and differentiate \( \frac{1}{\sin^{-1} x} \) using the chain rule, resulting in \( -\frac{(1-x^2)^{1/2}}{x} \).
04
Evaluate the New Limit
The new limit is \( \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{(1-x^2)^{1/2}}{x}} = \lim_{x \to 0^+} -\frac{1}{(1-x^2)^{1/2}} \). As \( x \to 0^+ \), \( (1-x^2)^{1/2} \to 1 \). Therefore, the limit simplifies to \( \lim_{x \to 0^+} -1 = -1 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hôpital's Rule
L'Hôpital's Rule is a valuable technique in calculus used to solve limits that are in one of the indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms arise when trying to directly substitute the limit value into a function's expression results in an undefined expression. The essence of L'Hôpital's Rule is to differentiate the numerator and the denominator separately. Using L'Hôpital's Rule:
- First, ensure the limit leads to an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
- Next, differentiate the numerator and denominator individually.
- Re-evaluate the limit with these new derivatives.
Indeterminate Forms
Indeterminate forms occur when direct substitution in a limit problem results in an undefined or ambiguous moment, making the limit non-trivial to solve. Typical forms include:
- \( 0/0 \)
- \( \infty/\infty \)
- \( \infty - \infty \)
- \( 0 \cdot \infty \)
Differentiation
Differentiation is the fundamental process of finding the derivative of a function. It represents how a function's value changes as its input changes, essentially providing the rate of change or slope of the function at any point. In applying L'Hôpital's Rule, differentiation plays a crucial role:
- The numerator \( \ln x \) differentiates to \( \frac{1}{x} \).
- Utilizing the chain rule on \( \frac{1}{\sin^{-1} x} \), differentiate to \( -\frac{(1-x^2)^{1/2}}{x} \).
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \( \sin^{-1} x \), are essential for solving trigonometry-related problems as they allow you to find the angle that corresponds to a given sine value. They follow certain behaviors:
- \( \sin^{-1} x \) approximates \( x \) for small values of \( x \).
- The derivative of \( \sin^{-1} x \) is \( \frac{1}{(1-x^2)^{1/2}} \).