Problem 42 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand the Problem

We need to find the limit of \(x^{x^2}\) as \(x\) approaches 0 from the positive side. This is an indeterminate form 0 raised to 0, so we need to use logarithmic transformation.

Step 2: Apply Logarithmic Transformation

Let's set \(y = x^{x^2}\) and take the natural logarithm of both sides to deal with the power more easily: \(\ln y = x^2 \ln x\). Now, as \(x \to 0^+\), evaluate the limit of \(x^2 \ln x\).

Step 3: Evaluate the Limit of ln Expression

Substitute the limit expression: find \(\lim_{x \to 0^+} x^2 \ln x\). This expression is of the form \(0 \cdot (-\infty)\), so perform a variable substitution to use L'Hôpital's Rule.

Step 4: Apply Variable Substitution

Let \(u = \ln x\), which implies that as \(x \to 0^+\), \(u \to -\infty\) and \(x = e^u\). Substitute into the limit: \(\lim_{u \to -\infty} \frac{u}{e^{-2u}}\).

Step 5: Use L'Hôpital's Rule

Now we have a form \(\frac{-\infty}{-\infty}\) suitable for L'Hôpital's Rule. Differentiate the numerator and denominator: \( \lim_{u \to -\infty} \frac{1}{-2e^{-2u}} = \lim_{u \to -\infty} \frac{-1}{2e^{-2u}}\).

Step 6: Simplify and Evaluate

As \(u \to -\infty\), the denominator \(2e^{-2u}\) \( \to 0^+\), making the numerator approach 0. Thus the limit is 0. This implies \(\ln y \to 0\), so \(y \to e^0 = 1\).

Step 7: Interpret the Result

Since \(y = e^{0} = 1\), the original limit \(\lim_{x \to 0^+} x^{x^2} = 1\).

Key Concepts

L'Hôpital's Rule

Understanding how L'Hôpital's Rule works can significantly simplify finding limits, especially when faced with indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It provides a way to transform a difficult limit problem into a simpler one by using derivatives.

The key idea is straightforward:

• Find an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Differentiate the numerator and the denominator separately—not their quotient.
• Re-evaluate the limit with these new derivatives.

When applying L'Hôpital's Rule, ensure that each differentiation maintains these indeterminate forms. This often means repeating the process until the limit can be easily solved. Remember, converting an expression into a form where L'Hôpital's Rule is applicable often involves clever substitutions or other transformations.

Logarithmic Transformation

The logarithmic transformation is a powerful technique to make difficult expressions more manageable. It comes in handy when dealing with complicated powers or when expressions involve exponential forms.

Here’s how it works:

• Take the logarithm of both sides of an equation to simplify exponentials to multiplications. This is crucial for handling terms like \(x^{x^2}\).
• Use properties of logarithms, such as \( \ln(x^a) = a \ln x\), to further simplify the limit expression.

For example, in our original limit problem, setting \(y = x^{x^2}\) allows us to transform the complex power into \( \ln y = x^2 \ln x \). This transformation lets us focus on finding the limit of the transformed expression first, which can then be exponentiated back to return to the original terms.

Indeterminate Forms

Indeterminate forms represent expressions where standard limit computation rules fall apart, typically appearing as \( \frac{0}{0} \), \(\infty - \infty\), or \(0^0\). In our problem, the objective is to evaluate \(x^{x^2} \) as \(x \to 0^+\), which naturally presents an indeterminate form \(0^0\).

Handling these forms requires special techniques:

• Transform the expression into a more analyzable form through algebraic manipulations or substitutions.
• Apply limits laws or rules, like L'Hôpital's Rule, which are specifically designed to tackle these forms.

Understanding when and how to manipulate indeterminate forms can switch an impossible-to-solve limit into an approachable simple calculation. Always first identify the type of indeterminacy and then choose the best mathematical tool to resolve it.