Problem 28
Question
In Exercises \(27-38,\) (a) describe the type of indeterminate form (if any) that is obtained by direct substitution. (b) Evaluate the limit, using L'Hôpital's Rule if necessary. (c) Use a graphing utility to graph the function and verify the result in part (b). \(\lim _{x \rightarrow 0^{+}} x^{3} \cot x\)
Step-by-Step Solution
Verified Answer
The limit of the function \(x^{3} \cdot \cot(x)\) as \(x\) approaches zero from the positive side is \(0\).
1Step 1: Direct Substitution and Identify Indeterminate Form
First, substitute \(x = 0^{+}\) directly into the expression \(x^{3} \cot(x)\). This gives \(0^{3} \cdot \cot(0^{+}) = 0 \cdot \infty\), which is an indeterminate form.
2Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that the limit of the quotient of two functions as \(x\) approaches a certain value is equal to the limit of the quotient of their derivatives, given that the original limit is an indeterminate form. Convert \( \cot(x)\) to \( \cos(x)/\sin(x)\) making the function \( f(x) = x^{3} \cdot \cos(x)/\sin(x) \). Separate it into two functions: \(f(x) = x^{3}\) and \( g(x) = \cos(x)/\sin(x)\). Then find the derivative of these two functions: \(f'(x) = 3x^{2}\) and \(g'(x) = -\sin(x)/\sin^{2}(x) - \cos^{2}(x)/\sin^{2}(x)\). Now, apply L'Hôpital's Rule to find the limit of their quotient, that is \(\lim _{x \rightarrow 0^+} \frac{f'(x)}{g'(x)}\).
3Step 3: Evaluate the limit
Given that the derivative functions are \(f'(x) = 3x^{2}\) and \(g'(x) = -\sin(x)/\sin^{2}(x) - \cos^{2}(x)/\sin^{2}(x)\), the limit as \(x\) approaches \(0\) is found to be \(\lim _{x \rightarrow 0^{+}} \frac{3x^{2}}{-\sin(x)/\sin^{2}(x) - \cos^{2}(x)/\sin^{2}(x)} = 0\), because the numerator is tending to 0 while the denominator isn't.
4Step 4: Verify the result graphically
Use a graphing utility to graph the function \(y = x^{3} \cdot \cos(x)/\sin(x)\). As \(x\) approaches \(0\) from the positive side on the graph, \(y\) approaches \(0\) which confirms the result of the limit evaluation in the previous step.
Key Concepts
Limits and ContinuityL'Hôpital's RuleGraphical Verification of Limits
Limits and Continuity
Understanding limits is a foundational concept in calculus that deals with approaching a certain point on a function. Continuity, on the other hand, describes a function that is unbroken or uninterrupted at every point within its domain.
When evaluating limits like the provided exercise, where we seek the value of \( \lim _{x \rightarrow 0^{+}} x^{3} \cot x \) as \( x \) approaches \( 0 \) from the positive side, we often encounter what are known as indeterminate forms. One common indeterminate form is \( 0 \times \infty \), obtained by directly substituting \( x = 0^{+} \) into the expression \( x^{3} \cot(x) \). This is problematic because it provides no clear numeric limit, and it's where continuity plays its part. A function that is continuous at \( x = 0^{+} \) would allow for direct substitution. However, when faced with an indeterminate form, we resort to additional strategies, such as L'Hôpital's Rule.
Limits and continuity are essential in calculus because they allow us to understand the behavior of functions at points that are not clearly defined, and they have applications across many areas within mathematics and the sciences.
When evaluating limits like the provided exercise, where we seek the value of \( \lim _{x \rightarrow 0^{+}} x^{3} \cot x \) as \( x \) approaches \( 0 \) from the positive side, we often encounter what are known as indeterminate forms. One common indeterminate form is \( 0 \times \infty \), obtained by directly substituting \( x = 0^{+} \) into the expression \( x^{3} \cot(x) \). This is problematic because it provides no clear numeric limit, and it's where continuity plays its part. A function that is continuous at \( x = 0^{+} \) would allow for direct substitution. However, when faced with an indeterminate form, we resort to additional strategies, such as L'Hôpital's Rule.
Limits and continuity are essential in calculus because they allow us to understand the behavior of functions at points that are not clearly defined, and they have applications across many areas within mathematics and the sciences.
L'Hôpital's Rule
L'Hôpital's Rule provides a method for evaluating limits that result in indeterminate forms such as \( 0/0 \) or \( \infty/\infty \). In the given exercise, we first recognize the indeterminate form \( 0 \times \infty \) and transform the expression to allow for the application of L'Hôpital's Rule. The cotangent function, \( \cot(x) \), is rewritten as \( \cos(x)/\sin(x) \) to achieve a fraction.
Upon differentiating the numerator and the denominator separately, the process to apply L'Hôpital's Rule can be initiated. It involves calculating the derivatives \( f'(x) = 3x^{2} \) and \( g'(x) = -\sin(x)/\sin^{2}(x) - \cos^{2}(x)/\sin^{2}(x) \) and then taking the limit of their quotient as \( x \) approaches \( 0^{+} \) which results in \( 0 \).
Using L'Hôpital's Rule correctly depends on ensuring that the requirements of the rule are met, such as the existence of derivatives and the proper indeterminate form. This rule greatly simplifies finding limits that would otherwise be difficult or impossible to evaluate directly.
Upon differentiating the numerator and the denominator separately, the process to apply L'Hôpital's Rule can be initiated. It involves calculating the derivatives \( f'(x) = 3x^{2} \) and \( g'(x) = -\sin(x)/\sin^{2}(x) - \cos^{2}(x)/\sin^{2}(x) \) and then taking the limit of their quotient as \( x \) approaches \( 0^{+} \) which results in \( 0 \).
Using L'Hôpital's Rule correctly depends on ensuring that the requirements of the rule are met, such as the existence of derivatives and the proper indeterminate form. This rule greatly simplifies finding limits that would otherwise be difficult or impossible to evaluate directly.
Graphical Verification of Limits
Graphical analysis is a powerful tool for verifying limits. After working through an indeterminate form algebraically, like in our exercise, graphing the function can provide a visual confirmation of the result.
Using a graphing utility, we would plot \( y = x^{3} \cdot \cos(x)/\sin(x) \) and observe the behavior as \( x \) approaches \( 0^{+} \) from the positive direction. If the graph approaches the y-value found analytically (in this case, 0), it serves as verification for our computed limit and helps us understand the function's behavior near the point of interest.
While the graphical verification is not a proof, it can build intuition and provide a visual explanation alongside the analytic approach. It is especially helpful in educational settings to enhance understanding and to verify analytical results in a more tangible way.
Using a graphing utility, we would plot \( y = x^{3} \cdot \cos(x)/\sin(x) \) and observe the behavior as \( x \) approaches \( 0^{+} \) from the positive direction. If the graph approaches the y-value found analytically (in this case, 0), it serves as verification for our computed limit and helps us understand the function's behavior near the point of interest.
While the graphical verification is not a proof, it can build intuition and provide a visual explanation alongside the analytic approach. It is especially helpful in educational settings to enhance understanding and to verify analytical results in a more tangible way.
Other exercises in this chapter
Problem 28
Find the integral involving secant and tangent. $$ \int \tan ^{5} 2 x \sec ^{2} 2 x d x $$
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Verify the integration formula. $$ \int u^{n} \cos u d u=u^{n} \sin u-n \int u^{n-1} \sin u d u $$
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Find the integral. $$ \int \frac{x^{3}+x+1}{x^{4}+2 x^{2}+1} d x $$
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Use substitution to find the integral. $$ \int \frac{e^{x}}{\left(e^{2 x}+1\right)\left(e^{x}-1\right)} d x $$
View solution