Problem 77 - University Calculus: Early Transcendentals

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.

Step 1: Understand the Function

The expression we need to evaluate is \( \lim_{x \rightarrow 0^{+}} x \ln x \). As \( x \rightarrow 0^{+} \), \( \ln x \rightarrow -\infty \), so initially this looks like an indeterminate form \( 0 \cdot (-\infty) \), which suggests using L'Hôpital’s Rule.

Step 2: Rewrite the Expression As a Quotient

To apply L'Hôpital’s Rule, rewrite the expression \( x \ln x \) as a fraction: \( \frac{\ln x}{1/x} \). This emphasizes the form \( \frac{-\infty}{\infty} \), suitable for L'Hôpital’s Rule.

Step 3: Apply L'Hôpital's Rule

Take the derivative of the numerator and the denominator: - Derivative of \( \ln x \) is \( \frac{1}{x} \).- Derivative of \( 1/x \) is \( -\frac{1}{x^2} \).Thus, \( \lim_{x \rightarrow 0^{+}} \frac{\ln x}{1/x} = \lim_{x \rightarrow 0^{+}} \frac{1/x}{-1/x^2} \). Simplify this to \( \lim_{x \rightarrow 0^{+}} (-x) \).

Step 4: Evaluate the Simplified Expression

Evaluate \( \lim_{x \rightarrow 0^{+}} (-x) \). As \( x \rightarrow 0^{+} \), \( -x \rightarrow 0 \). So, the expression evaluates to 0.

Step 5: Validate the Correct Option

Compare this evaluation with the given options:- Option (a) incorrectly evaluates the limit directly as zero without justification.- Option (b) claims the limit is \(-\infty\), which is not correct since proper use of L'Hôpital’s Rule gives 0.- Option (c) gives an incorrect value of \(-1\) and mixes improper limits.- Option (d) correctly evaluates the limit step by step using L'Hôpital’s Rule and concludes with 0.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to find limits involving indeterminate forms. It applies to fractions that tend to forms like \( \frac{0}{0} \) or \( \frac{\pm \infty}{\pm \infty} \). When you encounter these forms, you can differentiate the numerator and denominator independently and then take the limit again.

To use L'Hôpital’s Rule:

• Ensure the limit results in an indeterminate form, specifically \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Take the derivative of the numerator and the denominator.
• Evaluate the limit of this new fraction.
• Repeat the process if another indeterminate form arises.

For example, in the problem above, we turn the original indeterminate form \( 0 \cdot (-\infty) \) into \( \frac{-\infty}{\infty} \), by rewriting the product as a fraction, which then allows us to apply L'Hôpital’s Rule. This is crucial since it simplifies our work and provides a clear path to evaluating sometimes tricky limits.

Indeterminate Forms

Indeterminate forms are expressions that appear during limit calculations, where direct substitution leads to forms like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), among others. These forms suggest uncertainty in the limit value, prompting further analysis.

To handle indeterminate forms:

• Recognize whether the expression is indeterminate: common forms are \( 0/0 \) and \( \infty/\infty \).
• If \( 0 \cdot (-\infty) \) is present, like in this exercise, rewrite it as a fraction: \( \frac{\ln x}{1/x} \).
• Use algebraic manipulation or apply L'Hôpital’s Rule to simplify the expression into a determinate form, facilitating the limit evaluation.

Understanding this concept is vital as it informs you when standard limit calculations might not apply, nudging you towards applying more advanced techniques like L'Hôpital’s Rule.

Logarithmic Functions

Logarithmic functions play a central part in calculus, especially when dealing with products and compositions that involve exponential growth or decay. The logarithmic function \( \ln(x) \) has specific properties that are crucial when evaluating limits.

Key properties of logarithms include:

• As \( x \) approaches zero from the positive side, \( \ln(x) \) approaches \(-\infty \).
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \), a fact used in differentiating products involving \( \ln(x) \).

In our exercise, \( x \ln(x) \) initially gives an indeterminate form \( 0 \cdot (-\infty) \), necessitating the transformation into a fraction for application of L'Hôpital’s Rule. Recognizing this behavior is essential for manipulating and evaluating such limits effectively. Logarithms convert multiplicative relationships into additive ones, often simplifying complex expressions and paving the way for more straightforward computational strategies.