Problem 66 - Thomas Calculus in SI Units

Question

Find the limits $$\lim _{x \rightarrow 0^{+}} \sin x \cdot \ln x$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Recognize the Problem Structure

The function given is the product \( \sin x \cdot \ln x \). As \( x \to 0^{+} \), \( \sin x \to 0 \) and \( \ln x \to -\infty \). This presents an indeterminate form \( 0 \cdot (-\infty) \), which we need to evaluate.

Step 2: Rewrite the Expression for Analysis

Rewrite the limit expression to facilitate the use of L'Hôpital's Rule: \( \lim_{x \to 0^{+}} \sin x \cdot \ln x \) can be rewritten as \( \lim_{x \to 0^{+}} \frac{\ln x}{\csc x} \), where \( \csc x = \frac{1}{\sin x} \). So, our form is now \( \frac{-\infty}{\infty} \).

Step 3: Apply L'Hôpital's Rule

Since the expression is of the form \( \frac{-\infty}{\infty} \), apply L'Hôpital's Rule. Differentiate the numerator and the denominator separately: the derivative of \( \ln x \) is \( \frac{1}{x} \) and the derivative of \( \csc x \) is \( -\csc x \cot x \).

Step 4: Evaluate the New Limit

The new limit is now \( \lim_{x \to 0^{+}} \frac{1/x}{-\csc x \cot x} = \lim_{x \to 0^{+}} \frac{\sin x}{x(-\cot x)} \). Simplify the expression: \( \cot x = \frac{\cos x}{\sin x} \), thus \( \frac{\sin x}{-x \cot x} = \frac{\sin^2 x}{-x \cos x} \). As \( \sin x \approx x \) for \( x \to 0 \), we further simplify \( \frac{x^2}{-x \cos x} = \frac{x}{-\cos x} \).

Step 5: Compute the Simplified Limit

Simplify \( \lim_{x \to 0^{+}} \frac{x}{-\cos x} \). For \( x \to 0\), \( -\cos x \approx -1 \). Therefore, the limit becomes \( \lim_{x \to 0^{+}} -x = 0 \).

Step 6: Conclusion

The evaluation of the limit using simplification and approximations shows that \( \lim _{x ightarrow 0^{+}} \sin x \cdot \ln x = 0 \). The algebra confirms that the dominating factor as \( x \rightarrow 0^{+} \) is indeed the \( \sin x \) component.

Key Concepts

L'Hôpital's Rule

Often in calculus, you'll encounter complex limit problems that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). L'Hôpital's Rule is a powerful tool to resolve these situations. It allows us to differentiate the numerator and denominator separately rather than simplifying the fraction directly. This differentiation helps us obtain more definitive limit values.
To apply L'Hôpital's Rule, the function must be in an indeterminate form. For example, in the problem given, the limit \( \lim_{x \to 0^{+}} \frac{\ln x}{\csc x} \) was converted into a \( \frac{-\infty}{\infty} \) form. By differentiating \( \ln x \) and \( \csc x \), the problem simplifies significantly, allowing us to evaluate it with greater ease.
Remember, L'Hôpital's Rule applies only to these specific indeterminate forms. Therefore, recognize when a limit matches these patterns, and apply the rule methodically for a correct solution.

Indeterminate Forms

Indeterminate forms occur when substituting a limit into a function results in an undefined mathematical expression, like \( 0 \times (-\infty) \), \( \frac{0}{0} \), or \( \infty - \infty \). These expressions mean you can't immediately determine the value by direct substitution.
In our exercise, the product \( \sin x \cdot \ln x \) approaches \( 0 \times (-\infty) \) as \( x \) approaches zero from the right. This is an indeterminate form suggesting that straightforward evaluation isn't possible. Instead, you often need to rearrange or transform the equation to analyze it.
Clearing up the indeterminate form often involves mathematical tools like L'Hôpital's Rule, algebraic manipulation, or approximations. Recognizing and effectively dealing with them is crucial to solving advanced limit problems.

Trigonometric Limits

Trigonometric limits frequently appear in calculus, and understanding them is vital. These limits often require us to understand the behavior of trigonometric functions as they approach specific values, especially close to zero.
A key approximation used is that \( \sin x \approx x \) when \( x \) is near zero. This approximation become crucial in the process of simplifying expressions formed during limit analysis.
For example, as seen in the given solution, simplifying to \( \frac{x^2}{-x \cos x} \) and then reducing to \( \frac{x}{-\cos x} \) involved applying trigonometric limits for small \( x \), specifically using the behavior of \( \sin x \). Knowing these basic limit properties simplifies the process and ensures accurate results.