Chapter 8: Problem 12
Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} 3 x^{2} \csc ^{2} x $$
Short Answer
Expert verified
The limit is 3.
Step by step solution
01
Recognize the Indeterminate Form
To apply l'Hôpital's Rule, we need to determine if the limit \[\lim _{x \rightarrow 0} 3 x^{2} \csc ^{2} x\]is an indeterminate form. The function can be rewritten using the identity \(\csc x = \frac{1}{\sin x}\): \[3x^2 \csc^2 x = 3x^2 \frac{1}{\sin^2 x}\]which is equivalent to:\[\frac{3x^2}{\sin^2 x}\].As \(x\) approaches 0, both the numerator \(3x^2\) and the denominator \(\sin^2 x\) approach 0, producing the indeterminate form \(\frac{0}{0}\). Hence, we can apply l'Hôpital's Rule.
02
Differentiate the Numerator and Denominator
Using l'Hôpital's Rule, we differentiate the numerator and the denominator. For the numerator \(3x^2\), the derivative is:\[\frac{d}{dx}(3x^2) = 6x\]For the denominator, \(\sin^2 x\), the derivative is found using the chain rule:\[\frac{d}{dx}(\sin^2 x) = 2\sin x \cdot \cos x = \sin 2x\].
03
Apply l'Hôpital's Rule to the Derivatives
Apply l'Hôpital's Rule by taking the limit of the derivatives:\[\lim _{x \rightarrow 0} \frac{6x}{\sin 2x}\]Again, evaluate the limit. As \(x\) approaches 0, again both the numerator \(6x\) and the denominator \(\sin 2x\) approach 0, presenting another indeterminate form \(\frac{0}{0}\). We need to apply l'Hôpital's Rule again.
04
Differentiate the New Numerator and Denominator Again
Differentiate the new numerator and denominator. For \(6x\), the derivative is:\[\frac{d}{dx}(6x) = 6\]For \(\sin 2x\), the derivative using the chain rule is:\[\frac{d}{dx}(\sin 2x) = 2\cos 2x\].
05
Evaluate the Simplified Limit
Substitute in the new derivatives into the limit, yielding:\[\lim _{x \rightarrow 0} \frac{6}{2\cos 2x} = \lim _{x \rightarrow 0} \frac{3}{\cos 2x}\]Since \(\cos 2x\) approaches \(\cos 0 = 1\) as \(x\) approaches 0:\[\frac{3}{\cos 0} = 3\].The limit equals 3.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits in Calculus
Limits are a fundamental concept in calculus that help us understand the behavior of functions as they approach certain points. In the context of the given exercise, the goal is to find the limit of the function \(3x^2 \csc^2 x\) as \(x\) approaches 0.
Understanding limits are crucial because they form the basis of derivatives and integrals, key concepts in calculus. In simple terms, a limit examines what happens to a function's output (value) as the input (variable) gets closer to a particular point.
To find limits:
Understanding limits are crucial because they form the basis of derivatives and integrals, key concepts in calculus. In simple terms, a limit examines what happens to a function's output (value) as the input (variable) gets closer to a particular point.
To find limits:
- Identify the function and the point it is approaching.
- Check the behavior of the function around that point.
- Determine whether the limit exists or not.
Indeterminate Forms
Indeterminate forms often occur when evaluating limits, leading to expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms imply that just substituting values directly won't work, as they don't provide a clear result.
In this exercise, rewriting the function \(3x^2 \csc^2 x\) as \(\frac{3x^2}{\sin^2 x}\) reveals an indeterminate form \(\frac{0}{0}\) as \(x\) approaches 0. This means that both the numerator and denominator approach zero simultaneously, which doesn't hint toward any specific value for the overall expression.
When faced with indeterminate forms, solutions include:
In this exercise, rewriting the function \(3x^2 \csc^2 x\) as \(\frac{3x^2}{\sin^2 x}\) reveals an indeterminate form \(\frac{0}{0}\) as \(x\) approaches 0. This means that both the numerator and denominator approach zero simultaneously, which doesn't hint toward any specific value for the overall expression.
When faced with indeterminate forms, solutions include:
- Using algebraic manipulation to simplify the expression.
- Applying l'Hôpital's Rule to differentiate and resolve the limit.
Derivatives in Calculus
Derivatives are central to calculus and are used to measure how a function changes as its input changes. They provide the slope of a function at any point, which is the rate of change.
In this context, using derivatives is essential when applying l'Hôpital's Rule to resolve indeterminate forms. To apply the rule, differentiate the numerator and the denominator separately and again find the limit of the resulting expression.
Steps include:
In this context, using derivatives is essential when applying l'Hôpital's Rule to resolve indeterminate forms. To apply the rule, differentiate the numerator and the denominator separately and again find the limit of the resulting expression.
Steps include:
- Find the derivative of each part of the fraction separately.
- In this exercise, the derivative of \(3x^2\) is \(6x\), while the derivative of \(\sin^2 x\) requires the chain rule, resulting in \(\sin 2x\).
- If the limit is still indeterminate, further differentiate until it can be resolved.