Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow \pi / 2} \frac{\ln (\sin x)^{3}}{\frac{1}{2} \pi-x} $$

Short answer

The limit is 0.

Step-by-step solution

Check for Indeterminate Form

Substitute \(x = \frac{\pi}{2}\) into \(\frac{\ln(\sin^3 x)}{\frac{1}{2}\pi - x}\). We have \(\ln(\sin(\frac{\pi}{2}))^3 = \ln(1) = 0\) and \(\frac{1}{2}\pi - \frac{\pi}{2} = 0\). This gives the form \(\frac{0}{0}\), an indeterminate form where l'Hôpital's Rule can be applied.

Differentiate the Numerator

Differentiate the numerator, \(\ln(\sin^3 x)\), with respect to \(x\). Using the chain rule, \(\frac{d}{dx}[\ln(u)] = \frac{1}{u}\cdot \frac{du}{dx}\), where \(u = (\sin x)^3\). Thus, \(\frac{d}{dx}[\ln(\sin^3 x)] = \frac{3\sin^2 x \cdot \cos x}{\sin^3 x} = \frac{3\cos x}{\sin x}\).

Differentiate the Denominator

Differentiate the denominator, \(\frac{1}{2} \pi - x\), with respect to \(x\). The derivative is \(-1\).

Apply l'Hôpital's Rule

Apply l'Hôpital's Rule by taking the limit of the ratio of the derivatives: \[\lim _{x \rightarrow \pi/2} \frac{\frac{3 \cos x}{\sin x}}{-1} = \lim _{x \rightarrow \pi/2} -\frac{3 \cos x}{\sin x}\].

Evaluate the Limit

Evaluate the limit \(\lim _{x \rightarrow \pi/2} -\frac{3 \cos x}{\sin x}\). As \(x\) approaches \(\frac{\pi}{2}\), \(\cos x\) approaches 0, hence the entire expression evaluates to 0.

Key concepts

l'Hôpital's Rule

l'Hôpital's Rule is a powerful mathematical tool used in calculus to evaluate limits that initially result in indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule provides a systematic method to tackle these tricky limits by focusing on the derivatives of the numerator and denominator separately.
When you encounter a limit problem, the first step is to determine whether the expression falls into an indeterminate form. If so, you can apply l'Hôpital's Rule, which allows you to differentiate the numerator and the denominator and then take the limit of the resulting expression instead.

• The original limit problem: \(\lim_{x \to c} \frac{f(x)}{g(x)}\)
• If it leads to \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), differentiate: \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\)

It's important to check the form of the expression before using the rule; otherwise, the rule wouldn't be valid or necessary. Here in this problem, the indeterminate form \(\frac{0}{0}\) was correctly identified, making l'Hôpital's Rule applicable.

Indeterminate Form

An indeterminate form in calculus is an expression in which direct substitution during limit evaluation results in a form that is ambiguous or undefined. The most common indeterminate forms are \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). These forms indicate that neither the numerator nor the denominator has a prevailing influence on the result, making the limit hard to predict.

To solve limits that result in an indeterminate form, various methods are used, one of which is l'Hôpital's Rule. Before using l'Hôpital’s Rule, it's crucial to confirm that the expression indeed results in an indeterminate form. This can be verified by substituting the limiting value directly into the function.

• Check the expression by substitution.
• Identify if it's \(0/0\) or another indeterminate form.
• Apply appropriate techniques like l'Hôpital’s Rule.

In our example, substitution led to \(\frac{0}{0}\), confirming that it's an indeterminate form, thereby justifying the use of l'Hôpital's Rule.

Chain Rule

The chain rule is an essential technique in differentiation. It is particularly useful when finding the derivative of composite functions, where one function is nested inside another. In a composite function \( f(g(x)) \), the chain rule helps differentiate by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

The chain rule formula is given by:

• \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

In this exercise, you have \(\ln((\sin x)^3)\) which requires the chain rule:

• Differentiate \(\ln u\) where \(u = (\sin x)^3\).
• Derivative: \(\frac{1}{u} \cdot \frac{du}{dx}\) where \(\frac{du}{dx} = 3\sin^2 x \cdot \cos x\).
• The result is \(\frac{3\cos x}{\sin x}\).

By applying the chain rule, complex derivatives break into more manageable parts, aiding in the use of l'Hôpital's Rule.

Differentiation

Differentiation is a core concept in calculus that involves finding the derivative of a function. The derivative represents how a function changes as its input changes, essentially showing the rate of change or slope of the function at any given point.

There are basic rules of differentiation such as the power rule, product rule, quotient rule, and indeed, the chain rule. Each rule provides a method for systematically finding the derivative of different forms of functions.

• For a function \(f(x)\), its derivative is denoted as \(f'(x)\) or \(\frac{d}{dx}[f(x)]\).
• With differentiation, you can find slopes of tangent lines or rates of change.

In this example, both the numerator and denominator of the original fraction were differentiated separately to obtain \(\frac{3\cos x}{\sin x}\) and \(-1\) respectively. This step was crucial for applying l'Hôpital's Rule. Differentiation is key to refining the indeterminate form into a solvable limit.