Problem 26 - Essential Calculus Early Transcendentals

Question

\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow 0^{+}} \sin x \ln x$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Form of the Limit

We need to find \(\lim _{x \rightarrow 0^{+}} \sin x \ln x\). As \(x\) approaches 0 from the positive side, \(\ln x\) approaches \(-\infty\) and \(\sin x\) approaches 0. This forms an indeterminate product of the form \(0 \times (-\infty)\). To resolve this, we must rewrite it as a fraction.

Step 2: Re-express Product as a Fraction

To apply L'Hospital's Rule, rewrite the product \(\sin x \ln x\) as a fraction. We can write it as \(\frac{\ln x}{1/\sin x}\), which becomes \(\frac{-\ln x}{csc x}\). As \(x \rightarrow 0^{+}\), \(-\ln x\) approaches \(\infty\) and \(csc x\) (which is \(1/\sin x\)) approaches \(\infty\), forming an indeterminate \(\frac{\infty}{\infty}\) type of limit.

Step 3: Apply l'Hospital's Rule

Since the limit is of the indeterminate form \(\frac{\infty}{\infty}\), we can apply l'Hospital's Rule, which allows us to find the limit of the derivatives of the numerator and the denominator:\[\lim _{x \rightarrow 0^{+}} \frac{-\ln x}{csc x} = \lim _{x \rightarrow 0^{+}} \frac{d}{dx}(-\ln x)}{\frac{d}{dx}(csc x)}.\]Compute the derivative of \(-\ln x\) as \(-\frac{1}{x}\) and the derivative of \(csc x\) as \(-csc x \cdot cot x\). Thus, the limit becomes:\[\lim _{x \rightarrow 0^{+}} \frac{-\frac{1}{x}}{-csc x \cdot cot x} = \lim _{x \rightarrow 0^{+}} \frac{1/x}{csc x \cdot cot x}.\]

Step 4: Simplify the Expression and Find the Limit

Simplify the expression \(\frac{1/x}{csc x \cdot cot x}\):\[\lim _{x \rightarrow 0^{+}} \frac{1}{x csc x \cdot cot x} = \lim _{x \rightarrow 0^{+}} \frac{1}{x \cdot \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x}} = \lim _{x \rightarrow 0^{+}} \frac{\sin^2 x}{x \cdot \cos x}.\]As \(x \rightarrow 0\), \(\sin^2 x \approx x^2\) and \(\cos x \approx 1\), simplifying the limit gives:\[\lim _{x \rightarrow 0^{+}} \frac{x^2}{x} = \lim _{x \rightarrow 0^{+}} x = 0.\] So the limit is 0.

Key Concepts

limit evaluation

Limit evaluation is a fundamental concept in calculus used to find the value a function approaches as the input approaches a certain point. In the exercise, we look at the limit \( \lim_{x \rightarrow 0^{+}} \sin x \ln x \). Evaluating limits can sometimes lead to straightforward expressions but often require subtle transformations to reveal their true behavior.

To properly evaluate this limit, recognize the behavior of \( \ln x \) and \( \sin x \) as \( x \) approaches zero from the right:

• \( \ln x \) tends to \(-\infty\).
• \( \sin x \) tends to 0.

These observations show us that we are dealing with an indeterminate form. This means direct substitution isn't feasible.

Instead, creative manipulation is required, such as rewriting the expression or applying specific calculus techniques like L'Hospital's Rule. This helps us transform the limit into a more workable form, where known calculus principles can be applied to reach a resolution.

indeterminate forms

Indeterminate forms arise when substituting values into a limit results in expressions like \(0/0\), \(\infty/\infty\), or \(0 \cdot \infty\). They signal a need for additional analysis to find the limit accurately.

In the exercise, substituting \( x \rightarrow 0^{+} \) into \( \sin x \ln x \) gives us \(0 \cdot (-\infty)\), an indeterminate form. This specific form shows both a dominating zero factor and a negatively infinite term, making the expression impossible to evaluate directly. In such scenarios, calculus techniques come into play to rescue us.

We address this by converting the problematic product into a quotable form:

• Rewrite \( \sin x \ln x \) as \(\frac{\ln x}{1/\sin x}\).
• This creates another indeterminate form, \(\frac{\infty}{\infty}\).

This form allows us to apply L'Hospital's Rule to evaluate the limit without getting caught in the intricacies of the original expression.

calculus techniques

In calculus, employing the right techniques is vital for overcoming challenges posed by complicated expressions. One powerful tool is L'Hospital's Rule, used specifically for resolving indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

With the exercise, after converting \( \sin x \ln x \) into \( \frac{\ln x}{1/\sin x} \), we utilize L'Hospital's Rule. This rule suggests taking derivatives of the numerator and the denominator until a determinate form appears.

Here's how it unfolds:

• Derive the expression \(-\ln x / (\csc x)\) by differentiating:
• Numerator derivative: \(-1/x\)
• Denominator derivative: \( -\csc x \cdot \cot x\)

The limit simplifies eventually to \( \lim_{x \rightarrow 0^{+}} \frac{\sin^2 x}{x \cdot \cos x}\). Simplifying further, this becomes \( \lim_{x \rightarrow 0^{+}} x \), leading to the conclusion that the limit is 0.

Applying L'Hospital's Rule reveals the true nature of a limit that initially seemed indeterminate, showcasing the power and versatility of calculus techniques.