Problem 29 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Rewriting the Limit for L'Hospital's Rule

The initial limit is not in an immediate indeterminate form that allows the direct application of l'Hospital's Rule. We begin by rewriting the expression: \( \sqrt{x} \ln x = \frac{\ln x}{\frac{1}{\sqrt{x}}} \). As \( x \to 0^+ \), \( \ln x \to -\infty \) and \( \frac{1}{\sqrt{x}} \to \infty \). This gives us an indeterminate form of type \( \frac{-\infty}{\infty} \).

Step 2: Apply L'Hospital's Rule

Now that we have an indeterminate form, we can apply l'Hospital's Rule. This involves differentiating the numerator and the denominator separately.The derivative of the numerator \( \ln x \) is \( \frac{1}{x} \).The derivative of the denominator \( \frac{1}{\sqrt{x}} \) is \( -\frac{1}{2}x^{-3/2} = -\frac{1}{2\sqrt{x^3}} \).Thus, we take the limit:\[ \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{\sqrt{x}}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{2\sqrt{x^3}}} \].

Step 3: Simplifying the Derivatives' Quotient

Simplify the expression by multiplying by the reciprocal of the denominator:\[ \frac{\frac{1}{x}}{-\frac{1}{2\sqrt{x^3}}} = \frac{1}{x} \times \left(-2\sqrt{x^3}\right) = \frac{-2\sqrt{x^3}}{x} = -2x^{1/2} \].

Step 4: Evaluate the Limit

Now we take the limit of \(-2x^{1/2}\) as \(x\) approaches 0 from the positive side:\[ \lim_{x \to 0^+} -2x^{1/2} = -2 \times 0 = 0 \].

Key Concepts

Understanding Indeterminate Forms

Indeterminate forms arise in calculus when evaluating limits, and they aren't immediately obvious like some straightforward limits. These forms can look confusing because they suggest multiple possibilities for a result. When you first encounter a limit problem, such as \( \lim _{x \rightarrow 0^{+}} \sqrt{x} \ln x \), it might not directly fit a simple pattern. However, you can reformulate the expression to reveal an indeterminate form.

Common types of indeterminate forms are \( \frac{0}{0} \), \( \frac{-\infty}{\infty} \), and \( 0 \cdot \infty \), among others. In our exercise, rewriting the expression \( \sqrt{x} \ln x \) as \( \frac{\ln x}{\frac{1}{\sqrt{x}}} \) results in the form \( \frac{-\infty}{\infty} \), which is one of these classic indeterminate examples, allowing us to apply l'Hospital's Rule. Understanding how to identify and manipulate these indeterminate forms is central to solving limits using this rule.

Essentials of Limits in Calculus

Limits are a fundamental concept in calculus, describing the behavior of a function as it approaches a particular point. The limit helps us understand the behavior and trends of the function near points where the exact value might not be directly available or clearly defined.

When we take a limit like \( \lim_{x \to 0^+} \sqrt{x} \ln x \), we are essentially trying to pinpoint what the product will approach as \( x \) gets infinitely close to \( 0 \) from the positive side. Limits can be approached from either direction, but in this case, we specifically look at the positive direction because \( \sqrt{x} \) requires \( x \) to be non-negative to maintain real number results.

By applying limits, we not only talk about the values approaching but also the tendencies or directions, which are crucial in many calculus applications such as continuity, derivatives, and integral evaluations.

The Role of Derivatives in L'Hospital's Rule

Derivatives stick closely with the operation of limits, especially when using l'Hospital's Rule. This rule is a way to handle indeterminate forms by transforming the limit into a potential simpler form. You do this by differentiating the numerator and the denominator separately.

In our example, the function \( \ln x \) has a derivative of \( \frac{1}{x} \), and \( \frac{1}{\sqrt{x}} \) gives \( -\frac{1}{2}x^{-3/2} \) as its derivative. The mechanics involve taking these derivatives, replacing the original expressions, and then evaluating the new limit: \[ \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{\sqrt{x}}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{2\sqrt{x^3}}} \].

Derivatives transform complex, sometimes awkward forms into more manageable ones that can be evaluated directly or further simplified. They show the rate of change, and through l'Hospital's Rule, allow us to determine the behavior of functions as they expand or contract around specific points.