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
Recognize the Indeterminate Form
The limit given is \( \lim _{x \rightarrow 0^{+}}\left(\sin ^{-1} x\right)(\ln x) \). As \(x \to 0^+\), \(\sin^{-1}(x) \to \sin^{-1}(0) = 0\) and \(\ln(x) \to -\infty\). This creates an indeterminate form of \(0 \cdot (-\infty)\). To apply L'Hospital's Rule, we should first express the expression in a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form.
02
Rewrite as a Quotient
To rewrite the product as a quotient, we use the property that \(a \cdot b = \frac{a}{1/b}\). Thus, \((\sin^{-1} x)(\ln x)\) can be rewritten as: \[ \lim _{x \rightarrow 0^{+}} \frac{\ln x}{1/\sin^{-1} x} \] Now we have a \( \frac{-\infty}{\infty} \) form, suitable for L'Hospital's Rule.
03
Apply L'Hospital's Rule
L'Hospital's Rule states that if \( \frac{f(x)}{g(x)}\) is an indeterminate form, then: \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]if the limit on the right side exists.Here, \(f(x) = \ln x\) and \(g(x) = 1/\sin^{-1}x\). Find the derivatives:\(f'(x) = \frac{1}{x}\) and \(g(x) = \frac{1}{\sin^{-1} x}\) thus \(g'(x) = \frac{-1}{(\sin^{-1} x)^2 \sqrt{1-x^2}}\).Now apply L'Hospital's Rule: \[ \lim _{x \rightarrow 0^{+}} \frac{1/x}{-1/(\sin^{-1} x)^2 \sqrt{1-x^2}} \].
04
Simplify the Limit Expression
The expression simplifies as follows: \[ \lim _{x \rightarrow 0^{+}} \frac{\sin^{-1} x \cdot \sqrt{1-x^2}}{-x} \]Since \(\sin^{-1}(x)\) approx becomes \(x\) for small \(x\), and \(\sqrt{1-x^2}\) approaches \(1\), simplifying further the expression gives us \[ \lim_{x \rightarrow 0^+} \frac{x \cdot 1}{-x} = \lim_{x \rightarrow 0^+} -1 = -1. \]
05
Conclusion
The limit of the function \(\lim _{x \rightarrow 0^{+}}\left(\sin ^{-1} x\right)(\ln x)\) evaluates to -1 using L'Hospital's Rule after rewriting the expression to fit the rule and simplifying.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hospital's Rule
L'Hospital's Rule is a fundamental tool in calculus used to evaluate indeterminate forms like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). This rule can streamline the process of finding limits where usual algebraic simplifications are not possible. Let's say we have a limit of the form \( \frac{f(x)}{g(x)} \) as \( x \to c \). If both \( f(x) \) and \( g(x) \) approach 0 or \( \infty \) as \( x \to c \), L'Hospital's Rule states to take the derivatives of \( f \) and \( g \), then re-evaluate the limit as follows:
- \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \)
Indeterminate Forms
In calculus, indeterminate forms arise when a limit is not immediately obvious because it doesn't result in a clear finite number. These forms include expressions like \( \frac{0}{0}, \frac{\infty}{\infty}, 0 \cdot \infty, \infty - \infty, \) among others.Consider the expression \( (\sin^{-1} x)(\ln x) \) as \( x \to 0^+ \), which is a product that forms \( 0 \cdot (-\infty) \). Since this isn't a direct form for L'Hospital's Rule, we transform it into a quotient by using algebraic manipulations.
- \( ab = \frac{a}{1/b} \)
Inverse Trigonometric Functions
Inverse trigonometric functions, like \( \sin^{-1} x \), are crucial in calculus as they help in finding the angle whose sine is \( x \). Important properties of \( \sin^{-1} x \) include:
- \( \sin^{-1} x \to 0 \) as \( x \to 0 \)
- \( \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \)
Logarithmic Functions
Logarithmic functions, denoted as \( \ln x \), are the inverses of exponential functions and are widely used in calculus to simplify expressions and solve problems. Some key properties include:
- \( \ln x \to -\infty \) as \( x \to 0^+ \)
- \( \frac{d}{dx}(\ln x) = \frac{1}{x} \)