Problem 14 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) Evaluate the following limits. To do so rigorously, it is useful to apply L'Hôpital's rule (Appendix F). Otherwise, use a calculator to guess the answers. i. \(\lim _{x \rightarrow \infty} \frac{\sqrt{x}}{\ln x}\) ii. \(\lim _{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}}\) (b) Which grows faster as \(x \rightarrow \infty, \ln x\) or \(\sqrt{x}\) ?

Step-by-Step Solution

Answer

(a)i: \( \infty \) (a)ii: 0 (b): \( \sqrt{x} \) grows faster than \( \ln x \).

Step 1: Solve (a)i: Apply L'Hopital's Rule

To apply L'Hopital's rule, take the derivative of the numerator and denominator separately. The derivative of \( \sqrt{x} \) is \( 0.5x^{-0.5} \) and the derivative of \( \ln x \) is \( x^{-1} \). We then take the limit of the ratio of these two derivatives. \( \lim _{x \rightarrow \infty} \frac{0.5x^{-0.5}}{x^{-1}}=\lim _{x \rightarrow \infty} 0.5x^{0.5} \) which equals to \( \infty \).

Step 2: Solve (a)ii: Apply L'Hopital's Rule

Similarly, we differentiate the numerator and denominator separately for the limit. The derivative of \( \ln x \) is \( x^{-1} \) and \( \sqrt{x} \) is \( 0.5x^{-0.5} \). Taking the limit of the ratio of these two derivatives gives \( \lim _{x \rightarrow \infty} \frac{x^{-1}}{0.5x^{-0.5}}=\lim _{x \rightarrow \infty} \frac{2}{x^{0.5}} \), and as \( x \rightarrow \infty \), this approaches 0.

Step 3: Solve (b): Compare Growth Rates

From (a)i, we discovered that as \( x \rightarrow \infty, \sqrt{x} \) grows faster than \( \ln x \), as the limit of their ratio was infinite. Thus, we can say that \( \sqrt{x} \) grows faster as \( x \rightarrow \infty \) compared to \( \ln x \).

Key Concepts

Limits of Functions

Understanding the limits of functions is essential for grasping the behavior of functions as the input gets infinitely large or small, or approaches a particular value. In calculus, the limit of a function at a certain point refers to the value that the function approaches as its input (or the variable) approaches that point. It's like predicting the trend of a function as you zoom in closer and closer on its graph.

For instance, evaluating the limit of a function when the variable approaches infinity, as in our exercise \( \lim _{x \rightarrow \infty} \frac{\sqrt{x}}{\ln x} \) and \( \lim _{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} \), is akin to finding out what the function's output will be if we continue to follow its pattern much further along the x-axis. Calculus provides us with tools, such as L'Hôpital's rule, to evaluate these limits when they are not immediately obvious, which is crucial in many fields, including physics and engineering.

Derivatives in Calculus

The derivative represents the rate at which a function's output changes with respect to changes in its input. In simple terms, it's like the 'speedometer' of a function, giving you the speed at any point in time. In the realm of calculus, derivatives serve as the cornerstone for understanding and predicting the behavior of functions. They are particularly useful when functions have indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), where direct substitution fails to provide a limit.

In the exercise above, we use derivatives to apply L'Hôpital's Rule when evaluating the limits. This entails taking the derivative of the top and bottom functions separately and then finding the limit of the new ratio. Through L'Hôpital's rule, we simplify complex limit problems into more solvable ones, allowing us to make precise predictions and comparisons about how functions grow and change.

Growth Rate Comparison

When comparing the growth rates of two functions, we're essentially asking which function gets larger faster as the input grows. This concept is vital when analyzing efficiency in algorithms, population growth in biology, or the interest rates in finance. In our exercise, we compared the growth of \( \ln x \) and \( \sqrt{x} \) as \( x \rightarrow \infty \).

By taking the ratio of these functions and determining the limit, we can identify which one eventually becomes dominant. In step (b), based on the previous limits calculated, we observed that \( \sqrt{x} \) grows faster than \( \ln x \) because, as \( x \) grows large, the ratio \( \frac{\sqrt{x}}{\ln x} \) increases without bound. This method of using limits to compare growth rates provides a systematic approach to establish relationships between different rates at which quantities grow, which is a key process in mathematical modeling and analysis.