Problem 97
Question
Find the limits in Exercises \(97-104\) $$ \lim _{x \rightarrow 0} \frac{\sin ^{-1} 5 x}{x} $$
Step-by-Step Solution
Verified Answer
The limit is 5.
1Step 1: Recognize the Form
The given limit is of the form \( \frac{f(x)}{g(x)} \) as \( x \to 0 \), specifically \( \lim_{x \to 0} \frac{\sin^{-1}(5x)}{x} \). Since both the numerator \( \sin^{-1}(5x) \) and the denominator \( x \) approach 0 as \( x \to 0 \), we can apply L'Hospital’s Rule, which is used for 0/0 indeterminate forms.
2Step 2: Apply L'Hospital's Rule
L'Hospital's Rule states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) yields an indeterminate form \( \frac{0}{0} \), then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the limit on the right-hand side exists. Thus, we differentiate the numerator and denominator.
3Step 3: Differentiate the numerator
The derivative of \( \sin^{-1}(5x) \) with respect to \( x \) is \( \frac{5}{\sqrt{1-(5x)^2}} \). This follows from the formula for the derivative of \( \sin^{-1}(u) \), which is \( \frac{1}{\sqrt{1-u^2}} \), multiplied by the derivative of \( u \), where \( u = 5x \).
4Step 4: Differentiate the denominator
The derivative of \( x \) with respect to \( x \) is simply 1.
5Step 5: Take the new limit
Substitute the derivatives back into the limit to get \( \lim_{x \to 0} \frac{\frac{5}{\sqrt{1-(5x)^2}}}{1} = \lim_{x \to 0} \frac{5}{\sqrt{1-(5x)^2}} \).
6Step 6: Evaluate the limit
Now, evaluate \( \lim_{x \to 0} \frac{5}{\sqrt{1-(5x)^2}} \). As \( x \to 0 \), \( (5x)^2 \to 0 \), and thus the expression becomes \( \frac{5}{\sqrt{1-0}} = 5 \). The limit evaluates to 5.
Key Concepts
Limit evaluationIndeterminate formsInverse trigonometric functions
Limit evaluation
Limit evaluation is a fundamental concept in calculus and is used to determine the value that a function approaches as the input approaches a certain point. It involves analyzing functions as they get indefinitely close to a particular input value.
In the original exercise, we look at the problem \[\lim _{x \rightarrow 0} \frac{\sin ^{-1} 5 x}{x} \]To solve this, we recognize it as evaluating a specific type of limit problem often encountered around indeterminate forms or functions involving trigonometry.
There are several techniques to evaluate limits:
In the original exercise, we look at the problem \[\lim _{x \rightarrow 0} \frac{\sin ^{-1} 5 x}{x} \]To solve this, we recognize it as evaluating a specific type of limit problem often encountered around indeterminate forms or functions involving trigonometry.
There are several techniques to evaluate limits:
- Direct substitution: Plugging the value into the function directly, when applicable.
- Factoring: Simplifying complex expressions by factoring to cancel terms.
- L'Hospital's Rule: Used when limits result in indeterminate forms.
Indeterminate forms
Indeterminate forms are expressions in calculus where a direct limit computation doesn't yield a clear answer. When attempting to evaluate limits, it's common to encounter indeterminate forms like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\) among others.
In our exercise, as \(x\) approaches 0, both the numerator \(\sin^{-1}(5x)\) and the denominator \(x\) approach 0, creating a \(\frac{0}{0}\) indeterminate form.
This is where L'Hospital's Rule can be particularly useful. It allows us to handle these indeterminate forms by taking derivatives:
In our exercise, as \(x\) approaches 0, both the numerator \(\sin^{-1}(5x)\) and the denominator \(x\) approach 0, creating a \(\frac{0}{0}\) indeterminate form.
This is where L'Hospital's Rule can be particularly useful. It allows us to handle these indeterminate forms by taking derivatives:
- Differentiate the numerator: The derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1-x^2}}\). Applying the chain rule for \(\sin^{-1}(5x)\) gives us: \(\frac{5}{\sqrt{1-(5x)^2}}\).
- Differentiate the denominator: The derivative of \(x\) is 1.
Inverse trigonometric functions
Inverse trigonometric functions, such as \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\), are the inverse operations of the basic trigonometric functions. They allow us to find angles based on trigonometric values. For instance, \(\sin^{-1}(x)\) represents the angle whose sine is \(x\).
The derivative of inverse trigonometric functions is critical in calculus, especially when evaluating limits involving these functions.
For the derivative of \(\sin^{-1}(x)\), it's given by:\[\frac{1}{\sqrt{1-x^2}}\]Applying chain rule, specifically with \(\sin^{-1}(5x)\):
The derivative of inverse trigonometric functions is critical in calculus, especially when evaluating limits involving these functions.
For the derivative of \(\sin^{-1}(x)\), it's given by:\[\frac{1}{\sqrt{1-x^2}}\]Applying chain rule, specifically with \(\sin^{-1}(5x)\):
- The chain rule states: If \(y = f(g(x))\), then \(y' = f'(g(x)) \cdot g'(x)\).
- Here, \(g(x) = 5x\), which derives to 5, then multiply by the derivative of \(\sin^{-1}(u)\) where \(u=5x\).
- This yields \(\frac{5}{\sqrt{1-(5x)^2}}\).
Other exercises in this chapter
Problem 95
Evaluate the integrals. \begin{equation}\int_{2}^{4} x^{2 x}(1+\ln x) d x\end{equation}
View solution Problem 96
Evaluate the integrals in Exercises \(85-94\) $$ \int_{-\sqrt{3}}^{1 / \sqrt{3}} \frac{\cos \left(\tan ^{-1} 3 x\right)}{1+9 x^{2}} d x $$
View solution Problem 97
Evaluate the integrals. \begin{equation}\int 3 x^{\sqrt{3}} d x\end{equation}
View solution Problem 98
Find the limits in Exercises \(97-104\) $$ \lim _{x \rightarrow 1^{+}} \frac{\sqrt{x^{2}-1}}{\sec ^{-1} x} $$
View solution