Problem 56 - University Calculus: Early Transcendentals

Question

Find the limits. $$\lim _{x \rightarrow \infty} x^{1 / 12 x}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Rewrite the Expression

The given expression is \( x^{1/12x} \). Let's rewrite it using a more convenient form to analyze its behavior as \( x \to \infty \): \( \exp\left(\frac{\log(x)}{12x}\right) \).

Step 2: Apply Limit Inside the Exponent

We need to find the limit of the exponent, \( \lim_{x \to \infty} \frac{\log(x)}{12x} \). Recognize this as an indeterminate form \( \frac{\infty}{\infty} \), appropriate for L'Hôpital's Rule.

Step 3: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator separately: the derivative of \( \log(x) \) is \( \frac{1}{x} \), and the derivative of \( 12x \) is 12. Apply the rule: \[ \lim_{x \to \infty} \frac{\log(x)}{12x} = \lim_{x \to \infty} \frac{1/x}{12} = \lim_{x \to \infty} \frac{1}{12x} \]

Step 4: Evaluate the Simplified Limit

Now evaluate: \( \lim_{x \to \infty} \frac{1}{12x} = 0 \). Since \( 12x \to \infty \), the fraction approaches zero.

Step 5: Evaluate the Original Limit

Substitute the limit result: \( \lim_{x \to \infty} \exp\left(\frac{\log(x)}{12x}\right) = \exp(0) = 1 \).

Key Concepts

L'Hôpital's Rule and its Application

L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits that present indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In this exercise, we encountered an indeterminate form, \( \frac{\infty}{\infty} \), which makes it suitable for L'Hôpital's Rule. Here is how it works:

• Differentiation: Differentiate the numerator and denominator of the fraction separately.
• Continuity Check: After differentiation, re-evaluate the limit.
• Repeat if Necessary: If the new expression is still indeterminate, apply L'Hôpital's Rule again.

The approach is useful because it simplifies complex limits into more manageable forms. In this case, applying the rule to the function \( \frac{\log(x)}{12x} \) simplified it to \( \frac{1}{12x} \), eventually leading to a limit of zero.

Understanding Indeterminate Forms

Indeterminate forms arise in calculus when evaluating certain limits where the outcome is not immediately obvious. Common indeterminate forms include:

• \( \frac{0}{0} \): The zero over zero form often occurs in limits involving polynomial functions.
• \( \frac{\infty}{\infty} \): Both the numerator and denominator tend towards infinity, suggesting a need for further analysis, such as via L'Hôpital's Rule.
• \( \infty - \infty \): A difference that is unpredictable due to balancing infinite quantities.

Recognizing these forms within a problem is essential, as it indicates that direct substitution isn’t sufficient. In our problem, the limit \( \frac{\log(x)}{12x} \) exhibited the \( \frac{\infty}{\infty} \) form, requiring us to use L'Hôpital's Rule to find a more straightforward solution.

Exploring Logarithmic Functions

Logarithmic functions, represented as \( \log(x) \), are the inverses of exponential functions and play a significant role in calculus, especially in limits and differentiation. Understanding their behavior helps tackle a range of calculus problems:

• Growth Characteristics: \( \log(x) \) grows very slowly compared to polynomial or exponential functions. This property gives insight when \( x \to \infty \).
• Differentiation: The derivative of \( \log(x) \) is \( \frac{1}{x} \). This result is critical when applying L'Hôpital's Rule to functions involving logarithms.
• Application: Logarithmic transformations are often used to simplify expressions or transform multiplicatively complex functions into additive, more digestible ones.

In this specific exercise, rewriting \( x^{1/12x} \) using \( \log(x) \) provided a format suitable for limit evaluation using calculus techniques, enabling a deeper analysis into how the expression behaves as \( x \to \infty \).