Problem 30 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0^{+}} x^{2} \ln x $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Recognize the indeterminate form

The expression \( x^{2} \ln x \) becomes problematic as \( x \rightarrow 0^{+} \) because \( \ln x \rightarrow -\infty \) and \( x^2 \rightarrow 0 \). This results in the indeterminate form \(0 \times (-\infty)\). We can rewrite it to apply l'Hospital's Rule.

Step 2: Rewrite the expression

Rewrite the expression as a quotient to use l'Hospital's Rule: \( x^2 \ln x = \frac{\ln x}{x^{-2}} \). Now, as \( x \rightarrow 0^{+} \), we have \( \frac{-\infty}{\infty} \), which is an indeterminate form suitable for l'Hospital's Rule.

Step 3: Differentiate the numerator and denominator

Differentiate the numerator and the denominator separately: the derivative of \( \ln x \) is \( \frac{1}{x} \) and the derivative of \( x^{-2} \) is \( -2x^{-3} \). Now we have the limit \( \lim_{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-2x^{-3}} \).

Step 4: Simplify and find the limit

Simplify the expression: \( \frac{\frac{1}{x}}{-2x^{-3}} = \frac{1}{x} \times \frac{-1}{2} x^3 = \frac{-x^2}{2} \). Taking the limit as \( x \rightarrow 0^{+} \), we see that \( x^2 \rightarrow 0 \). Thus, \( \lim_{x \rightarrow 0^{+}} \frac{-x^2}{2} = 0 \).

Key Concepts

Indeterminate Forms

When working with limits in calculus, you may encounter expressions that do not clearly tend towards a specific value. These are known as indeterminate forms. The name comes from the uncertainty these forms present in determining a limit directly. Indeterminate forms arise in various situations, commonly seen as:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \times \infty \)
• \( \infty - \infty \)
• \( 0^{0} \)
• \( \infty^{0} \)
• \( 1^{\infty} \)

These forms are "indeterminate" because they can potentially resolve in multiple ways depending on the function's behavior. For example, \( x^2 \ln x \) as \( x \rightarrow 0^{+} \) results in \( 0 \times (-\infty) \), meaning that as \( x \) approaches zero from the positive side, one part of the expression shrinks to zero while another grows without bound in the negative direction. To resolve such forms and find limits, techniques such as rewriting them or using l'Hospital's Rule can be invaluable.

Limits

The concept of limits is fundamental in calculus and helps us understand the behavior of functions as inputs approach a particular value. It answers the question of what value a function f(x) approaches as x gets closer to some number, potentially even when x never actually equals that number.In the exercise, you need to evaluate the limit \( \lim_{x \rightarrow 0^{+}} x^2 \ln x \). Here, by observing \( x^2 \) heading towards zero and \( \ln x \) diving towards \(-\infty\), we're led into an indeterminate form \( 0 \times (-\infty) \). By cleverly rewriting it into \( \frac{\ln x}{x^{-2}} \), we can examine a different indeterminate scenario, \( \frac{-\infty}{\infty} \), that is suitable for l'Hospital's Rule.Understanding limits can provide deep insights into how functions behave and help in calculating derivatives and integrals. Limits are used to define continuity, understand trends or end-behaviors, and are foundational for the notion of convergence in sequences and series.

Calculus

Calculus is the mathematical study of continuous change and consists of two main branches: differential calculus and integral calculus. Differential calculus focuses on rates of change and slopes of curves, leading to the concept of derivatives. Integral calculus deals with accumulation of quantities and areas under or between curves, introducing integrals.In our limit problem, calculus plays a vital role by employing derivatives to navigate through indeterminate forms. For instance, l'Hospital's Rule, a pivotal tool from differential calculus, becomes incredibly useful when simple substitution in a limit results in a form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule guides us to differentiate the numerator and the denominator separately, transforming the limit into an expression that can be resolved more straightforwardly.Employing calculus techniques allows us to manage complex limits, discover properties of functions, and solve real-world problems where change and accumulation are intrinsic. Its learning opens the door to understanding mathematical patterns that govern motion, growth, decay, and more.