Problem 7 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0} x^{2} \ln x $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Form of the Limit

First, we need to analyze the expression \( x^2 \ln x \) as \( x \to 0 \). Substituting \( x = 0 \) directly, we encounter the expression \( 0^2 \ln(0) \). While \( 0^2 = 0 \), \( \ln(0) \) tends to negative infinity. This represents an indeterminate form of \( 0 \cdot (-\infty) \).

Step 2: Transform the Expression

To resolve the indeterminate form, we can rewrite the expression as a fraction: \( x^2 \ln x = \frac{\ln x}{1/x^2} \). This changes the problem to finding \( \lim_{x \to 0} \frac{\ln x}{1/x^2} \), which is in \( \frac{-\infty}{\infty} \) form.

Step 3: Apply l'Hôpital's Rule

The expression is in a \( \frac{-\infty}{\infty} \) form, so we apply l'Hôpital's Rule, which involves taking derivatives of the numerator and denominator. Differentiate: \( (\ln x)' = \frac{1}{x} \) and \( \left(\frac{1}{x^2}\right)' = -\frac{2}{x^3} \). The limit becomes \( \lim_{x \to 0} \frac{1/x}{-2/x^3} = \lim_{x \to 0} \frac{-x^2}{2} \).

Step 4: Evaluate the Simplified Limit

Now, evaluate the simpler expression: \( \lim_{x \to 0} \frac{-x^2}{2} \). As \( x \to 0 \), \( -x^2 \to 0 \), so the expression becomes \( \frac{0}{2} = 0 \).

Step 5: Conclude the Solution

Therefore, the original limit \( \lim_{x \to 0} x^2 \ln x \) evaluates to 0.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms often arise during limit evaluation. These forms occur when the limit process leads to expressions that are not immediately clear, such as \(0/0\) or \(\infty - \infty\). For example, in our given problem, substituting directly into \(x^2 \ln x\) as \(x \to 0\) poses a challenge. While \(x^2\) approaches 0, \(\ln x\) heads towards negative infinity, resulting in the indeterminate form \(0 \cdot (-\infty)\). Since such expressions do not provide usable information about the limit, strategies to transform them into a more determinate form, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), are often necessary. This transformation allows us to apply powerful tools like l'Hôpital's Rule to resolve the limit.

Limit Evaluation

Limit evaluation is a fundamental technique in calculus, enabling us to understand the behavior of functions as they approach a specific point. In our exercise, we need to evaluate the limit \(\lim_{x \to 0} x^2 \ln x\). Direct substitution results in an indeterminate form, prompting us to transform it into a fraction: \(\frac{\ln x}{1/x^2}\). This reshaped expression simplifies the job by putting it into a known form: \(\frac{-\infty}{\infty}\). Once transformed, we can apply methods like l'Hôpital's Rule. The goal is to keep simplifying the expression until an easily computable limit emerges. In this case, by carefully applying derivatives, the expression simplifies to \(\lim_{x \to 0} \frac{-x^2}{2}\), which directly evaluates to 0.

Derivatives

Derivatives are crucial in the application of l'Hôpital's Rule, which is a technique for evaluating limits involving indeterminate forms. The process involves finding the derivatives of the numerator and the denominator of a troublesome fraction. For the limit \(\lim_{x \to 0} \frac{\ln x}{1/x^2}\), we differentiate the numerator and the denominator separately. The derivative of \(\ln x\) is \(\frac{1}{x}\), and the derivative of \(\frac{1}{x^2}\) is \(-\frac{2}{x^3}\). By applying these derivatives, the limit is transformed into \(\lim_{x \to 0} \frac{1/x}{-2/x^3}\), simplifying further to \(\lim_{x \to 0} \frac{-x^2}{2}\). This systematic approach allows us to tackle complex limits by leveraging the properties of derivatives.