Problem 46 - Calculus for Biology and Medicine

Question

(a) Show that $$\lim _{x \rightarrow \infty} \frac{\ln x}{x}=0$$ (b) Use your result in (a) to show that, for any \(c>0\), $$c x \geq \ln x$$ for sufficiently large \(x\).46. (a) Show that $$\lim _{x \rightarrow \infty} \frac{\ln x}{x}=0$$ (b) Use your result in (a) to show that, for any \(c>0\), $$c x \geq \ln x$$ for sufficiently large \(x\). (c) Use your result in (b) to show that, for any \(p>0\), $$x^{p} e^{-x} \leq e^{-x / 2}$$ provided that \(x\) is sufficiently large. (d) Use your result in (c) to show that, for any \(p>0\), $$\int_{0}^{\infty} x^{p} e^{-x} d x$$ is convergent.

Step-by-Step Solution

Answer

(a) \( \lim_{x \to \infty} \frac{\ln x}{x} = 0 \). (b) \( cx \geq \ln x \) for large \( x \). (c) \( x^p e^{-x} \leq e^{-x/2} \) for large \( x \). (d) The integral converges.

Step 1: Understanding the limit of the ratio

To show \( \lim_{x \rightarrow \infty} \frac{\ln x}{x} = 0 \), we can use L'Hôpital's Rule. Both the numerator and denominator approach infinity as \( x \to \infty \). L'Hôpital's Rule applies here since we have an indeterminate form \( \frac{\infty}{\infty} \). Differentiating the numerator and the denominator gives: \( \lim_{x \rightarrow \infty} \frac{\ln x}{x} = \lim_{x \rightarrow \infty} \frac{(\ln x)'}{x'} = \lim_{x \rightarrow \infty} \frac{1/x}{1} = \lim_{x \rightarrow \infty} \frac{1}{x} = 0 \). Therefore, the limit is \( 0 \).

Step 2: Comparing the functions for large x

Given the result from part (a), for any \( c > 0 \), you are to show that \( c x \geq \ln x \) for sufficiently large \( x \). As \( x \) grows, \( \frac{\ln x}{x} \to 0 \), meaning \( \ln x \) becomes negligible compared to \( c x \). Thus, eventually \( \ln x \ll c x \), so \( c x \geq \ln x \) holds true for sufficiently large \( x \).

Step 3: Relating exponentials with polynomials

Using \( c x \geq \ln x \) for large \( x \), we want to show \( x^{p} e^{-x} \leq e^{-x/2} \) for \( p > 0 \). For large \( x \), the term \( x^p \) grows much slower than the exponential decay in \( e^{-x} \). By choosing a large enough \( x \), \( x^p \leq e^{x/2} \) can hold since the exponential function \( e^{x/2} \) grows faster. Multiplying both sides of the inequality by \( e^{-x} \), we have \( x^p e^{-x} \leq e^{-x/2} \).

Step 4: Checking the convergence of the integral

To show \( \int_{0}^{\infty} x^{p} e^{-x} dx \) is convergent, we use the result from part (c). For large \( x \), since \( x^{p} e^{-x} \leq e^{-x/2} \), the integral from some point to infinity will be bounded by the convergent integral \( \int_{0}^{\infty} e^{-x/2} dx \). As the smaller integral converges, and since both behave similarly, the integral \( \int_{0}^{\infty} x^p e^{-x} dx \) must also converge.

Key Concepts

L'Hôpital's Rule

When encountering limits in calculus, such as \( \lim _{x \rightarrow \infty} \frac{\ln x}{x} = 0 \), L'Hôpital's Rule is a powerful tool. This rule is applicable in cases where you get an indeterminate form like \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \). L'Hôpital's Rule suggests that you can find the limit by differentiating the numerator and the denominator separately.

In our problem, the numerator is \( \ln x \) and its derivative is \( 1/x \). The denominator is \( x \) with a derivative of \( 1 \). Applying L'Hôpital's Rule gives us the new limit \( \lim_{x \rightarrow \infty} \frac{1/x}{1} \) which simplifies to \( \lim_{x \rightarrow \infty} \frac{1}{x} = 0 \). Thus, the function \( \frac{\ln x}{x} \) indeed approaches 0 as \( x \) grows large.

This result emphasizes how dramatically faster the linear function \( x \) grows compared to the logarithmic function \( \ln x \) when \( x \) is very large.

Exponential Functions

Exponential functions are mathematical functions in the form \( e^x \), where \( e \) is Euler's number, approximately equal to 2.718. These functions grow rapidly and are fundamental in the study of calculus, especially in behavior analysis of different sorts of functions when \( x \) is large or approaches infinity.

In our exercise, we looked at the inequalities involving exponential functions. For sufficiently large \( x \), \( x^p \) becomes insignificant compared to \( e^x \). This means if \( p \) is any positive number, no matter how large \( x^p \) grows, \( e^x \) is eventually much larger.

The task showed that \( x^p e^{-x} \leq e^{-x/2} \) by demonstrating that as \( x \) increases, the exponential decay in the term \( e^{-x} \) overpowers the polynomial growth in \( x^p \). This illustrates the dominant nature of exponential functions over polynomial ones as \( x \) becomes large.

Integral Convergence

The concept of integral convergence is crucial in determining whether the total "area" under a curve is finite. For the integral \( \int_{0}^{\infty} x^p e^{-x} dx \), convergence means that the total area under the curve from 0 to infinity is a finite number.

To check this, we argued in earlier steps that \( x^p e^{-x} \leq e^{-x/2} \) for large \( x \). Since \( \int_{0}^{\infty} e^{-x/2} dx \) converges (because the exponential function \( e^{-x/2} \) rapidly diminishes to zero), by comparison, so does \( \int_{0}^{\infty} x^p e^{-x} dx \).

• Comparison tests are useful to show convergence by comparing with a function known to converge.
• The exponential function's rapid decrease ensures convergence even if \( x^p \) increases slowly.

Therefore, the behavior of \( e^{-x} \) guarantees that the integral of \( x^p e^{-x} \) converges despite the polynomial growth.