Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each limit. Be sure you have an indeterminate form before applying l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} 3 x^{2} \csc ^{2} x $$

Step-by-Step Solution

Answer

The limit is 3.

Step 1: 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.

Step 2: 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:\[\frac{d}{dx}(\sin^2 x) = 2\sin x \cdot \cos x = \sin 2x\].

Step 3: 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.

Step 4: 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:\[\frac{d}{dx}(\sin 2x) = 2\cos 2x\].

Step 5: 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.

Key Concepts

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:

• 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.

Limits that don't resolve directly can sometimes lead to indeterminate forms, which require special methods to evaluate, like l'Hôpital's Rule in this exercise.

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:

• Using algebraic manipulation to simplify the expression.
• Applying l'Hôpital's Rule to differentiate and resolve the limit.

Understanding these forms helps in determining when and how to apply l'Hôpital's Rule effectively to find limits.

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:

• 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.

The derivative helps simplify the process of evaluating complex limits, making it possible to solve problems that seem challenging at first glance.