Problem 358 - Calculus Volume 1

Question

Evaluate the limit. Evaluate the limit \(\lim _{x \rightarrow \infty} \frac{\ln x}{x^{k}}\).

Step-by-Step Solution

Answer

The limit evaluates to 0 as \( x \to \infty \).

Step 1: Identify the Limit Form

The given limit is \( \lim_{x \rightarrow \infty} \frac{\ln x}{x^{k}} \). As \( x \) approaches infinity, both the numerator and denominator increase, making it an indeterminate form \( \frac{\infty}{\infty} \). Therefore, we can use L'Hôpital's Rule to evaluate this limit.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that for limits of indeterminate forms \( \frac{\infty}{\infty} \), the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) if the latter limit exists. Here, \( f(x) = \ln x \) and \( g(x) = x^k \). First, find their derivatives: \( f'(x) = \frac{1}{x} \) and \( g'(x) = kx^{k-1} \).

Step 3: Compute the Limit of the Derivatives

Substitute the derivatives back into the limit: \[ \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{kx^{k-1}} = \lim_{x \rightarrow \infty} \frac{1}{kx^{k}}. \]

Step 4: Evaluate the Limit

As \( x \to \infty \), \( kx^k \to \infty \). Therefore, \( \frac{1}{kx^k} \to 0 \). Thus, the original limit evaluates to 0.

Key Concepts

Limits

In calculus, a limit is used to describe the value that a function approaches as the input approaches a specified point. Limits are fundamental in understanding the behavior of functions as inputs reach values that may not be explicitly defined within the function. Whether it's approaching infinity or zero, limits help in predicting the function's behavior without directly computing values.Here's how it works:

• Consider the function given in the exercise: \[ rac{\ln x}{x^k} \]
• We are asked to evaluate this function as \( x \) approaches infinity.
• This helps in understanding what happens to the function far along its graph.

The computed limit tells us about the long-term behavior of the function, providing valuable insight into functions that are not necessarily straightforward to evaluate at face value. Understanding limits can help in simplifying otherwise complex expressions, making them an essential part of calculus.

L'Hôpital's Rule

L'Hôpital's Rule comes to the rescue when evaluating limits that result in indeterminate forms. This rule is particularly useful in situations where directly substituting values into a limit results in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).To apply L'Hôpital's Rule effectively:

• Identify that the limit is in an indeterminate form.
• Differentiate both the numerator and the denominator separately.
• Evaluate the limit of the resulting fraction of derivatives.

In the given exercise, the expression \( \lim_{x \to \infty} \frac{\ln x}{x^k} \) presents an \( \frac{\infty}{\infty} \) form, perfect for L'Hôpital's Rule. By differentiating \( \ln x \) to get \( \frac{1}{x} \) and \( x^k \) to get \( kx^{k-1} \), we obtain a simpler function to evaluate, leading to a clearer understanding of the original limit expression.

Indeterminate Forms

Indeterminate forms occur in calculus when a limit calculation involves uncertain expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and others. These uncertainties mean that such expressions do not straightforwardly indicate a finite value.In this particular exercise:

• The form \( \frac{\infty}{\infty} \) arises when both the numerator and denominator tend towards infinity as \( x \to \infty \).
• This tells us that direct evaluation will not simply work, as the infinite nature of the components causes ambiguity.
• L'Hôpital's Rule is especially helpful, providing a method to resolve this indeterminacy by considering the behavior of the derivatives instead.

Recognizing and handling indeterminate forms is crucial because it sets apart calculus from basic algebra, offering a deeper toolbox for analysis when simple substitution fails.

Limit at Infinity

The concept of limits at infinity examines the behavior of a function as its independent variable approaches infinity. It's a look into the end-behavior of a function, determining tendencies and settling rates when plotted over a large domain.In the exercise:

• We observe what happens to \( \frac{\ln x}{x^k} \) as \( x \) increases without bound.
• Finding the limit at infinity allows us to discern whether the function stabilizes at a certain value, continues to grow, or diminishes.
• The solution demonstrates that, despite both numerator and denominator growing, the denominator outpaces the growth, tending the function towards zero.

By evaluating the limit at infinity, we derive that not every function merely grows larger indefinitely, but some indeed approach finitely levelling out to a specified value, commonly zero for such ratios.