Chapter 4: Problem 100
Use limit methods to determine which of the two given functions grows faster, or state that they have comparable growth rates. $$x^{2} \ln x ; x^{3}$$
Short Answer
Expert verified
Answer: The function \(x^3\) grows faster than \(x^2 \ln x\) as x approaches infinity.
Step by step solution
01
Set up the Ratio of the Functions
First, let's represent the first given function as \(f(x) = x^2 \ln x\) and the second function as \(g(x) = x^3\). Then take the ratio of the functions:
$$\frac{f(x)}{g(x)} = \frac{x^2 \ln x}{x^3}$$
02
Simplify the Ratio
Now, let's simplify the ratio by simplifying the exponents and canceling out common terms:
$$\frac{x^2 \ln x}{x^3} = \frac{\ln x}{x}$$
03
Use L'Hopital's Rule if Necessary
To find the limit of the simplified ratio \(\frac{\ln x}{x}\) as \(x\) approaches infinity, we can use L'Hopital's rule if the limit is in the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). However, in this case, the limit is not in the indeterminate form, so we cannot use L'Hopital's rule.
04
Evaluate the Limit
Calculate the limit of the simplified ratio as x approaches infinity:
$$\lim_{x \to \infty} \frac{\ln x}{x}$$
Since the natural logarithm function increases slower than a linear function when x approaches infinity, the limit is equal to zero:
$$\lim_{x \to \infty} \frac{\ln x}{x} = 0$$
05
Interpret the Result
According to the limit result, we can say that \(g(x) = x^3\) grows faster than \(f(x) = x^2 \ln x\) as the ratio of their growth rates approaches zero when x approaches infinity. So, the function \(x^3\) has a faster growth rate compared to \(x^2 \ln x\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Methods
Understanding limit methods is key when you want to analyze the behavior of functions as their inputs either get very large or very small. A limit represents a value that a function approaches as the input approaches some value.
For the function ratio \(\frac{x^2 \ln x}{x^3}\), we aim to find its limit as \(x\) approaches infinity. This can reveal which of the two original functions, \(x^2 \ln x\) or \(x^3\), increases faster. We started by simplifying the ratio, which is a common method, to make it easier to find the limit. In simplifying, we eliminated common terms, leaving us with \(\frac{\ln x}{x}\).
Next, we assess if the expression is an indeterminate form that would require special limit methods, such as L'Hopital's Rule. If not, we can often find limits by observing the behavior of the numerator and denominator independently as \(x\) becomes very large. For \(\frac{\ln x}{x}\), as \(x\) grows, \(\ln x\) grows much slower than \(x\), leading to the conclusion that the limit is 0.
For the function ratio \(\frac{x^2 \ln x}{x^3}\), we aim to find its limit as \(x\) approaches infinity. This can reveal which of the two original functions, \(x^2 \ln x\) or \(x^3\), increases faster. We started by simplifying the ratio, which is a common method, to make it easier to find the limit. In simplifying, we eliminated common terms, leaving us with \(\frac{\ln x}{x}\).
Next, we assess if the expression is an indeterminate form that would require special limit methods, such as L'Hopital's Rule. If not, we can often find limits by observing the behavior of the numerator and denominator independently as \(x\) becomes very large. For \(\frac{\ln x}{x}\), as \(x\) grows, \(\ln x\) grows much slower than \(x\), leading to the conclusion that the limit is 0.
L'Hopital's Rule
L'Hopital's Rule is an elegant way to solve specific limit problems involving indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When you encounter a limit that leads to one of these indeterminate forms, you can differentiate the numerator and the denominator separately and then take the limit again.
However, it's crucial to identify when L'Hopital's Rule is applicable. In our exercise, the limit of \(\frac{\ln x}{x}\) as \(x \to \infty\) does not result in an indeterminate form hence L'Hopital's cannot be used immediately. This rule is a powerful tool, but remember to check if the conditions for its use are met before applying it.
In general, it's recommended to first attempt to simplify the expression or find the limit through direct substitution or factoring before deploying L'Hopital's Rule.
However, it's crucial to identify when L'Hopital's Rule is applicable. In our exercise, the limit of \(\frac{\ln x}{x}\) as \(x \to \infty\) does not result in an indeterminate form hence L'Hopital's cannot be used immediately. This rule is a powerful tool, but remember to check if the conditions for its use are met before applying it.
In general, it's recommended to first attempt to simplify the expression or find the limit through direct substitution or factoring before deploying L'Hopital's Rule.
Natural Logarithm
The natural logarithm, represented as \(\ln x\), is a special kind of logarithm with the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. It's a fundamental concept in calculus because of its unique properties and its occurrence in various mathematical situations, especially in growth and decay problems.
The natural logarithm increases at a decreasing rate as its input grows larger. In other words, while \(\ln x\) keeps increasing as \(x\) increases, it does so more slowly compared to polynomial functions. In the context of our exercise, the slow growth rate of the natural logarithm compared to the polynomial function \(x\) helps us conclude that \(\ln x / x\) goes to zero as \(x\) approaches infinity, indicating that \(x^2 \ln x\) grows slower than \(x^3\).
The natural logarithm increases at a decreasing rate as its input grows larger. In other words, while \(\ln x\) keeps increasing as \(x\) increases, it does so more slowly compared to polynomial functions. In the context of our exercise, the slow growth rate of the natural logarithm compared to the polynomial function \(x\) helps us conclude that \(\ln x / x\) goes to zero as \(x\) approaches infinity, indicating that \(x^2 \ln x\) grows slower than \(x^3\).
Indeterminate Forms
Indeterminate forms occur when a limit presents an expression that is not immediately clear or defined, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms don't provide an obvious answer and suggest that further analysis is necessary to determine the limit.
Being able to identify indeterminate forms is critical, as they often indicate situations where L'Hopital's Rule can be applied to find the limit. Other common indeterminate forms include \(0 \times \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\).
In the problem we are addressing, the limit we needed to evaluate did not fall into the indeterminate form category. Therefore, we were able to determine the functions' growth rates by analyzing the behavior of the logarithmic function relative to the polynomial function directly.
Being able to identify indeterminate forms is critical, as they often indicate situations where L'Hopital's Rule can be applied to find the limit. Other common indeterminate forms include \(0 \times \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\).
In the problem we are addressing, the limit we needed to evaluate did not fall into the indeterminate form category. Therefore, we were able to determine the functions' growth rates by analyzing the behavior of the logarithmic function relative to the polynomial function directly.
