Problem 49 - Essential Calculus

Question

\(\begin{array}{llll}\text { Comparing } & \text { Functions } & \text { In Exercises } & \mathbf{4 5}-\mathbf{5 0}, & \text { use } & \text { L'Hôpital's }\end{array}\) Rule to determine the comparative rates of increase of the functions \(f(x)=x^{m}, \quad g(x)=e^{n x},\) and \(h(x)=(\ln x)^{n}\) where \(n>0, m>0,\) and \(x \rightarrow \infty\). \(\lim _{x \rightarrow \infty} \frac{(\ln x)^{n}}{x^{m}}\)

Step-by-Step Solution

Answer

The limit \(\lim _{x \rightarrow \infty} \frac{(\ln x)^{n}}{x^{m}} = 0\). Therefore, as \(x \rightarrow \infty\), function \(f(x)\) increases at a faster rate compared to function \(h(x)\).

Step 1: Identify the Form of the Limit

Firstly, identify the form of the limit. As \(x \rightarrow \infty\), \((\ln x)^n\) becomes ∞ and \(x^m\) also becomes ∞, hence the limit has the form ∞/∞.

Step 2: Apply L’Hôpital’s Rule

In such a case where the limit is of the form ∞/∞, L'Hopital's Rule can be applied. L'Hopital's Rule states that such a limit is equal to the limit of the ratio of their derivatives, if it exists. Therefore, differentiate the numerator and denominator separately.

Step 3: Differentiate the numerator and denominator

The derivative of \((\ln x)^n\) with respect to \(x\) is \(\frac{n(\ln x)^{n-1}}{x}\) and derivative of \(x^m\) with respect to \(x\) is \(mx^{m-1}\). The limit therefore transforms into \(\lim _{x \rightarrow \infty} \frac{n(\ln x)^{n-1}/x}{mx^{m-1}}\). This simplifies to \(\lim _{x \rightarrow \infty} \frac{n(\ln x)^{n-1}}{mx^m}\).

Step 4: Evaluate the Limit

The new limit still has the form ∞/∞, so we continue to implement L'Hôpital’s Rule until we arrive at a determinate form. The limit then evaluates to zero.

Key Concepts

Comparative Rates of Increase

When comparing functions that approach infinity, it's important to determine how fast each function grows. This helps understand which function "dominates" as the input grows larger. In the exercise, three functions were considered: \(f(x) = x^m\), \(g(x) = e^{nx}\), and \(h(x) = (\ln x)^n\). Each of these functions behaves differently at infinity. The task is to determine which function grows the fastest as \(x\) approaches infinity.

L'Hôpital's rule is a powerful tool for understanding these rates. Based on the comparison:

• Polynomials like \(x^m\) tend to grow slower than exponential functions such as \(e^{nx}\).
• Logarithmic functions, even when raised to a power like \((\ln x)^n\), grow the slowest at infinity.

By applying L'Hôpital's rule repeatedly, you can clearly see these relationships in mathematical terms.

Limits at Infinity

Limits at infinity explore the behavior of functions as \(x\) tends towards infinity or negative infinity. In our exercise, we want to find out what happens to \(\lim_{x \rightarrow \infty} \frac{(\ln x)^n}{x^m}\).

As \(x\), the functions \((\ln x)^n\) and \(x^m\) both approach infinity. This results in an indeterminate form \(\frac{\infty}{\infty}\). L'Hôpital's rule is specifically used in these situations to resolve such uncertainties by converting the limit into a simpler form.

By differentiating both the numerator and the denominator and taking the limit again, we get a clearer picture. Through multiple iterations, the derivative approach allows us to resolve the indeterminate form into a conclusion, which in this case shows that the limit evaluates to zero.

Differentiation

Differentiation is the process of finding the derivative of a function. A derivative represents the rate of change of a function with respect to a variable. In solving our exercise, we use differentiation to apply L'Hôpital's rule repeatedly.

When you differentiate \((\ln x)^n\), it becomes \(\frac{n (\ln x)^{n-1}}{x}\). This tells us how the logarithmic term changes as \(x\) changes. On the other hand, differentiating \(x^m\) yields \(mx^{m-1}\), indicating how the polynomial function changes as \(x\) increases.

By replacing the original limit with these derivatives, we transform the limit \(\lim_{x \rightarrow \infty} \frac{(\ln x)^n}{x^m}\) into \(\lim_{x \rightarrow \infty} \frac{n (\ln x)^{n-1}}{mx^m}\). This simplifies the expression and makes it easier to evaluate as we apply L'Hôpital’s rule repeatedly to resolve the problem.