Problem 77 - Thomas Calculus in SI Units

Question

Only one of these calculations is correct. Which one? Why are the others wrong? Give reasons for your answers. a. \(\lim _{x \rightarrow 0^{+}} x \ln x=0 \cdot(-\infty)=0\) b. \(\lim _{x \rightarrow 0^{+}} x \ln x=0 \cdot(-\infty)=-\infty\) c. \(\lim _{x \rightarrow 0^{+}} x \ln x=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{(1 / x)}=\frac{-\infty}{\infty}=-1\) d. \(\lim _{x \rightarrow 0^{+}} x \ln x=\lim _{x \rightarrow 0^{+}} \frac{\ln x}{(1 / x)}\) \(=\lim _{x \rightarrow 0^{+}} \frac{(1 / x)}{\left(-1 / x^{2}\right)}=\lim _{x \rightarrow 0^{+}}(-x)=0\)

Step-by-Step Solution

Answer

Option (d) is correct; the limit as \( x \to 0^{+} \) is 0.

Step 1: Evaluate Expression of Part (a)

For option (a), the expression is evaluated as \( x \to 0^{+} \). The form \( x \ln x = 0 \cdot (-\infty) \) is an indeterminate form and cannot be simply evaluated as 0 without proper analysis. Hence, the conclusion they make, zero, is incorrect.

Step 2: Evaluate Expression of Part (b)

Option (b) concludes that the expression \( x \ln x = 0 \cdot (-\infty) = -\infty \). Like option (a), this is an indeterminate form. Therefore, it is also incorrect to conclude directly that \( -\infty \) is the result.

Step 3: Evaluate Expression of Part (c)

In option (c), the expression is rewritten using L'Hôpital's Rule as \( \lim _{x \rightarrow 0^{+}} \frac{\ln x}{(1 / x)} \), which is \( \frac{-\infty}{\infty} \), leading them to conclude \(-1\). This calculation mistakenly assumes \(-1\) without correct derivatives application, making it incorrect.

Step 4: Evaluate Expression of Part (d)

In option (d), the same transformation is applied as in (c), \( \lim _{x \rightarrow 0^{+}} \frac{\ln x}{(1 / x)} \). Using L'Hôpital's Rule correctly, we find \( \lim _{x \rightarrow 0^{+}} \frac{1/x}{-1/x^2} = \lim _{x \rightarrow 0^{+}} (-x) = 0 \). Using L'Hôpital’s Rule accurately leads to the conclusion that the limit is 0. This calculation is correct.

Key Concepts

Indeterminate Forms

In calculus, when evaluating limits, we often encounter expressions that result in undefined or ambiguous values, known as indeterminate forms. These forms include expressions like \(0 \cdot \infty\), \(\frac{0}{0}\), or \(\frac{\infty}{\infty}\). The presence of an indeterminate form signals that we need to apply additional mathematical methods to find a precise limit.

In the given problem, both options (a) and (b) involve the expression \(x \ln x\) as \(x \to 0^{+}\), which take the form \(0 \cdot (-\infty)\). This is an indeterminate form, and we cannot directly conclude the value of the limit without further analysis. Indeterminate forms require techniques like algebraic manipulation or specific calculus rules to resolve them into determinable values.

L'Hôpital's Rule

When confronted with indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's Rule offers a pathway to evaluate limits. This rule states that if a limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) exists, it can be found by taking the derivative of the numerator and the derivative of the denominator, then re-evaluating the limit.

In this exercise, L'Hôpital’s Rule was applied to evaluate option (d). Given the transformation \(\lim _{x \rightarrow 0^{+}} \frac{\ln x}{1/x}\), the rule requires taking derivatives: the derivative of \(\ln x\) is \(1/x\) and the derivative of \(1/x\) is \(-1/x^2\). Simplifying \(\frac{1/x}{-1/x^2}\) results in \(-x\). As \(x \to 0^{+}\), \(-x\) approaches 0, revealing that the correct limit is 0.

Calculus Problem-Solving

Approaching calculus problems effectively often requires a structured method. Evaluate the form of the expression, determine if special rules apply, and systematically process with the appropriate techniques. In this exercise, understanding the type of limit and the initial form are crucial.

By recognizing that each form, particularly the presence of an indeterminate form, needs extra care, we avoid premature conclusions. Options (a) and (b) incorrectly concluded results without resolving the indeterminate nature, whereas option (c) wrongly applied L'Hôpital's Rule leading to an incorrect result. A proper problem-solving approach identifies these errors and corrects them by applying correct rules and transformations, as successfully done in option (d).

One-Sided Limits

One-sided limits focus on approaching a certain value from only one direction on the number line. In this case, the limit \(\lim_{x \to 0^{+}}\) denotes that \(x\) approaches 0 from the right, meaning it considers only positive values of \(x\) approaching zero.

This is important because the behavior of functions can differ dramatically when approaching from the left or the right. For the expression \(x \ln x\) as \(x \to 0^{+}\), values of \(x\) are positive, leading \(\ln x\) to negative infinity. Understanding one-sided limits helps in correctly analyzing and evaluating such expressions, particularly in cases involving logarithms where inputs must remain positive to avoid undefined values.