Problem 64
Question
a. Prove that \(\lim _{x \rightarrow \infty} \frac{x^{k}}{e^{x}}=0\) for every positive constant \(k\). This shows that the natural exponential function approaches infinity faster than any power function. b. Prove that \(\lim _{x \rightarrow \infty} \frac{\ln x}{x^{k}}=0\) for every positive constant \(k\). This shows that the natural logarithmic function approaches infinity slower than any power function.
Step-by-Step Solution
Verified Answer
In summary, we proved that \(\lim _{x \rightarrow \infty} \frac{x^{k}}{e^{x}}=0\) for every positive constant \(k\) and \(\lim _{x \rightarrow \infty} \frac{\ln x}{x^{k}}=0\) for every positive constant \(k\) using L'Hôpital's rule. This shows that the natural exponential function approaches infinity faster than any power function and the natural logarithmic function approaches infinity slower than any power function.
1Step 1: Use L'Hôpital's Rule
Since \(\lim_{x \rightarrow \infty} x^{k} = \infty\) and \(\lim_{x \rightarrow \infty} e^x = \infty\), we can apply L'Hôpital's rule, which states that if \(\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \frac{\infty}{\infty}\), then \(\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}\), if the latter limit exists.
2Step 2: Find the derivatives of the numerator and denominator
First, differentiate \(x^k\) with respect to \(x\), resulting in \(k\cdot x^{k-1}\). Then, differentiate \(e^x\) with respect to \(x\), resulting in \(e^x\).
3Step 3: Apply L'Hôpital's Rule to the limit
Using L'Hôpital's Rule, we will now find \(\lim_{x \rightarrow \infty} \frac{k\cdot x^{k-1}}{e^x}\).
4Step 4: Determine the new limit
Since \(\lim_{x \rightarrow \infty} k\cdot x^{k-1} = \infty\) and \(\lim_{x \rightarrow \infty} e^x = \infty\), we can continue applying L'Hôpital's Rule until the exponent on \(x\) in the numerator is \(0\). When this happens, we have \(\lim_{x \rightarrow \infty} \frac{k!}{e^x} = 0\), because the exponential function grows faster than the factorial function. Therefore, \(\lim _{x \rightarrow \infty} \frac{x^{k}}{e^{x}}=0\) for every positive constant \(k\).
b. Proving \(\lim _{x \rightarrow \infty} \frac{\ln x}{x^{k}}=0\) for every positive constant \(k\)
5Step 1: Use L'Hôpital's Rule
Since \(\lim_{x \rightarrow \infty} \ln x = \infty\) and \(\lim_{x \rightarrow \infty} x^{k} = \infty\) for every positive constant \(k\), we can apply L'Hôpital's rule.
6Step 2: Find the derivatives of the numerator and denominator
First, differentiate \(\ln x\) with respect to \(x\), resulting in \(\frac{1}{x}\). Then, differentiate \(x^k\) with respect to \(x\), resulting in \(k\cdot x^{k-1}\).
7Step 3: Apply L'Hôpital's Rule to the limit
Using L'Hôpital's Rule, we will now find \(\lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{k\cdot x^{k-1}}\).
8Step 4: Simplify the limit
The limit can be simplified as \(\lim_{x \rightarrow \infty} \frac{1}{k\cdot x^{k}}\).
9Step 5: Determine the new limit
Since \(\lim_{x \rightarrow \infty} \frac{1}{k\cdot x^{k}} = \frac{1}{\infty} = 0\), we have proved that \(\lim _{x \rightarrow \infty} \frac{\ln x}{x^{k}}=0\) for every positive constant \(k\).
Key Concepts
Limits of FunctionsExponential GrowthLogarithmic FunctionsPower Functions
Limits of Functions
Limits help us understand the behavior of functions as they approach specific points or infinity. It's about what happens to a function's value as the variable gets closer to a particular number or trends to infinity.
In calculus, limits can be used to describe the continuity or discontinuity of functions as well as the potential end behavior of graphs. For example, in the given exercise, we analyzed the limits as \(x\) tends towards infinity.
In calculus, limits can be used to describe the continuity or discontinuity of functions as well as the potential end behavior of graphs. For example, in the given exercise, we analyzed the limits as \(x\) tends towards infinity.
- The expression \(\lim _{x \rightarrow \infty} \frac{x^k}{e^x} = 0\) indicates that a power function's growth is outpaced by the exponential function as \(x\) increases.
- Similarly, \(\lim _{x \rightarrow \infty} \frac{\ln x}{x^k} = 0\) shows us that the logarithmic growth of a function is too slow to keep up with even the simplest quadratic or higher power functions as \(x\) increases.
Exponential Growth
Exponential functions express really rapid growth or decay. The most common exponential function \(e^x\), where \(e\) is the base of natural logarithms, showcases exponential growth perfectly.
In the problem, we saw that \(e^x\) increases faster than any power function \(x^k\). This means:
In the problem, we saw that \(e^x\) increases faster than any power function \(x^k\). This means:
- In a scenario where \(x\) keeps growing, \(e^x\) will rise faster than \(x^2\), \(x^3\), and so on.
- This property is incredibly powerful in mathematics and real-world applications, such as finance for compound interest calculations or biology for population growth models.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions. The natural logarithm denoted by \(\ln x\) gives the time required for any exponential growth to reach a certain level.
Logarithms grow very slowly, especially compared to power or exponential functions. In the exercise, we proved that \(\ln x\) increases much slower than \(x^k\) for any positive constant \(k\).
Logarithms grow very slowly, especially compared to power or exponential functions. In the exercise, we proved that \(\ln x\) increases much slower than \(x^k\) for any positive constant \(k\).
- Logarithmic functions are commonly used in calculations dealing with growth rates, such as the Richter scale for earthquake magnitudes or the pH scale in chemistry.
Power Functions
Power functions take the form \(x^k\), where \(k\) is any real number. They can model a variety of growth patterns depending on the exponent \(k\).
When \(k\) is positive, the function \(x^k\) denotes polynomial growth, which is significant but predictable. In contrast to exponential functions, power functions grow considerably slower as demonstrated in the exercise.
When \(k\) is positive, the function \(x^k\) denotes polynomial growth, which is significant but predictable. In contrast to exponential functions, power functions grow considerably slower as demonstrated in the exercise.
- We saw that \(x^k\) could not compete with \(e^x\), showing how power functions are restricted in their rapid growth.
- However, they outpace logarithmic functions like \(\ln x\).
Other exercises in this chapter
Problem 63
Spending Health-care spending per person by the private sector comprising payments by individuals, corporations, and their insurance companies is approximated b
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Find the derivative of the function. $$ f(x)=\cos ^{-1}(\sin 2 x) $$
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Find the derivative of the function. $$ g(x)=\frac{2}{\sqrt{3^{x}+1}} $$
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