Problem 43 - Calculus for Biology and Medicine

Question

(a) Show that $$ \lim _{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}}=0 $$ (b) Use your result in (a) to show that $$ 2 \ln x \leq \sqrt{x} $$ for sufficiently large \(x .\) Use a graphing calculator to determine just how large \(x\) must be for \((7.17)\) to hold. (c) Use your result in (b) to show that $$ \int_{0}^{\infty} e^{-\sqrt{x}} d x $$ converges. Use a graphing calculator to sketch the function \(f(x)=e^{-\sqrt{x}}\) together with its comparison function(s), and use your graph to explain how you showed that the integral in (7.18) is convergent.

Step-by-Step Solution

Answer

(a) The limit is 0. (b) For \(x > 100\), \(2 \ln x \leq \sqrt{x}\). (c) Integral converges using comparison test.

Step 1: Calculating the Limit

We start by evaluating the limit \( \lim _{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} \). As \( x \rightarrow \infty \), both \( \ln x \) and \( \sqrt{x} \rightarrow \infty \). To use L'Hôpital's Rule, differentiate the numerator and denominator: - Derivative of \( \ln x \) is \( \frac{1}{x} \).- Derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).Apply L'Hôpital's Rule: \[ \lim _{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} = \lim _{x \rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} = \lim _{x \rightarrow \infty} \frac{2\sqrt{x}}{x} = \lim _{x \rightarrow \infty} \frac{2}{\sqrt{x}} = 0. \]

Step 2: Demonstrating Inequality

We use the result from Step 1 to show \( 2 \ln x \leq \sqrt{x} \) for sufficiently large \( x \). As \( \lim _{x \rightarrow \infty} \frac{\ln x}{\sqrt{x}} = 0 \), it implies that for very large \( x \), \( \ln x \) is much smaller than \( \sqrt{x} \). Therefore, \( 2 \ln x \leq \sqrt{x} \) holds true when \( x \) is large enough.To determine when the inequality holds, use a graphing calculator to plot \( 2 \ln x \) and \( \sqrt{x} \). The curves will show that they intersect when \( x \) is sufficiently large. Typically, values beyond \( x > 100 \) or more will illustrate this inequality.

Step 3: Confirming Convergence of the Integral

Based on Step 2's result, if \( 2 \ln x \leq \sqrt{x} \) for large \( x \), \( e^{-\sqrt{x}} \leq e^{-2\ln x} = \frac{1}{x^2} \). The function \( \frac{1}{x^2} \), starting from a certain \( x \), convergences as its integral from \( x=1 \) to infinity is finite. Hence, \( \int_{0}^{\infty} e^{-\sqrt{x}} dx \) must also converge.Using a graphing calculator, you can sketch \( f(x)=e^{-\sqrt{x}} \) and superimpose \( \frac{1}{x^2} \) to see that \( e^{-\sqrt{x}} \) decreases faster, demonstrating convergence by comparison.

Key Concepts

L'Hôpital's Rule

When dealing with limits that result in indeterminate forms, such as \( \frac{\infty}{\infty} \), L'Hôpital's Rule provides an invaluable tool. It states that if you have a limit that results in such an indeterminate form, you can take the derivatives of the numerator and the denominator, and re-evaluate the limit. This rule applies only under specific conditions: both the original functions \( f(x) \) and \( g(x) \) must be differentiable, and \( g'(x) eq 0 \) near the point of interest. In the original exercise, we needed to evaluate the limit \( \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} \). Direct substitution leads to an indeterminate form \( \frac{\infty}{\infty} \). Utilizing L'Hôpital's Rule, we differentiate the numerator and the denominator: the derivative of \( \ln x \) is \( \frac{1}{x} \), and the derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \). By applying the rule, we reformulate the limit as follows: \[ \lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{2\sqrt{x}}} = \lim_{x \to \infty} \frac{2\sqrt{x}}{x} = \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0. \]Thus, L'Hôpital's Rule allows us to simplify complex indeterminate limits, bringing clarity and precision to our calculations.

Convergence of Improper Integrals

Improper integrals extend the concept of an integral to unbounded intervals or integrands. To determine if these integrals converge or diverge, we often use comparison tests or direct evaluation. Convergence means the integral results in a finite value, while divergence implies it does not.In the exercise, we aimed to show the convergence of \( \int_{0}^{\infty} e^{-\sqrt{x}} \, dx \). Using a result from earlier in our solution process such as \( 2 \ln x \leq \sqrt{x} \) for large \( x \) gave us an inequality: \( e^{-\sqrt{x}} \leq \frac{1}{x^2} \) when \( x \) is sufficiently large. Given that the integral of \( \frac{1}{x^2} \) from \( x = 1 \) to infinity converges, we can use this comparison to conclude that \( \int_{0}^{\infty} e^{-\sqrt{x}} \, dx \) must also converge. Comparison and limit comparison tests are critical in examining improper integrals, where convergence can be demonstrated by relating to a known function with a similar or faster rate of decreasing.

Logarithmic and Radical Functions

Logarithmic and radical functions appear frequently in calculus, coming with their own unique behavior as \( x \) becomes very large. Understanding their growth rates is key, as illustrated in the inequality \( 2 \ln x \leq \sqrt{x} \) derived from the exercise.Logarithmic functions, like \( \ln x \), grow slowly compared to polynomial functions or radicals. As \( x \) increases, the rate of increase of \( \ln x \) becomes progressively smaller, never becoming infinite.Radical functions, like \( \sqrt{x} \), grow faster than logarithmic functions but slower than polynomials of higher degree. This growth rate relationship was exploited in the exercise to show that from a certain point onward, \( \sqrt{x} \) heavily outweighs \( 2 \ln x \). Using a graphing tool, students can visually compare these functions and verify that \( 2 \ln x \leq \sqrt{x} \) holds for sufficiently large \( x \). Graphs of functions provide an additional dimension of understanding, making abstract inequality arguments tangible and easier to comprehend.