Question

Use limit methods to determine which of the two given functions grows faster, or state that they have comparable growth rates. $$\ln x ; \ln (\ln x)$$

Short answer

Answer: The function \(\ln x\) grows faster than the function \(\ln (\ln x)\).

Step-by-step solution

Write down the given functions

The given functions are \(\ln x\) and \(\ln (\ln x)\). We will compare their growth rates as x approaches infinity.

Create a ratio representing the growth rates

To compare the growth rates, we create a ratio of the two functions, such that: $$R(x) = \frac{\ln x}{\ln (\ln x)}$$

Determine the limit of the ratio as x approaches infinity

We now find the limit of the ratio as x approaches infinity: $$\lim_{x\to\infty} R(x) = \lim_{x\to\infty} \frac{\ln x}{\ln (\ln x)}$$

Use L'Hôpital's Rule

Since the limit of the ratio as x approaches infinity is in the indeterminate form \(\frac{\infty}{\infty}\), we can use L'Hôpital's Rule. This rule states: $$\lim_{x\to\infty} \frac{f(x)}{g(x)}= \lim_{x\to\infty} \frac{f'(x)}{g'(x)}$$ Let \(f(x)=\ln x\) and \(g(x)=\ln (\ln x)\). So, $$f'(x)=\frac{1}{x} \quad\text{and}\quad g'(x)=\frac{1}{x\ln x}$$ Now, apply L'Hôpital's Rule to find the limit of the ratio: $$\lim_{x\to\infty} R(x) = \lim_{x\to\infty}\frac{\frac{1}{x}}{\frac{1}{x\ln x}}$$

Simplify the expression and evaluate the limit

After simplifying the expression, we get: $$\lim_{x\to\infty} R(x) = \lim_{x\to\infty} \frac{x\ln x}{x}= \lim_{x\to\infty} \ln x$$ Since \(\lim_{x\to\infty} \ln x = \infty\), we can conclude the following:

Conclusion

The function \(\ln x\) grows faster than the function \(\ln (\ln x)\).

Key concepts

Growth Rate Comparison

When comparing growth rates in calculus, we often want to determine which function increases more quickly as we move along the x-axis. To do so, a common approach is to form a ratio of the two functions, and then examine the behavior of this ratio as the variable, typically x, approaches infinity.

In our example, the functions \(\ln x\) and \(\ln (\ln x)\) are being compared through the ratio \(\frac{\ln x}{\ln (\ln x)}\). We want to determine how this ratio behaves as x grows larger and larger—does it approach zero, infinity, or remains constant? If the ratio approaches infinity, we can say that the numerator, the function \(\ln x\), grows faster than the denominator, \(\ln (\ln x)\). Conversely, if the ratio approaches zero, the denominator grows faster.

L'Hôpital's Rule

L'Hôpital's Rule is an invaluable tool in calculus for resolving situations where we encounter indeterminate forms such as \(\frac{\infty}{\infty}\) or \(\frac{0}{0}\). When we apply this rule, we differentiate the numerator and the denominator independently and then take the limit of the new ratio.

In our problem, when we attempt to find the limit of the ratio as x approaches infinity, we face the indeterminate form \(\frac{\infty}{\infty}\). This is where L'Hôpital's Rule comes into play. By differentiating the numerator \(\ln x\) and the denominator \(\ln (\ln x)\), and then considering the limit of the resulting new ratio, we can often find a more straightforward path to the answer.

Indeterminate Forms

Indeterminate forms occur in calculus when evaluating limits results in an expression that is not immediately obvious in determining the limit's behavior. Common examples of indeterminate forms include \(\frac{\infty}{\infty}\), \(\frac{0}{0}\), \(\infty - \infty\), 0 \times \infty, \(\infty^{0}\), and more.

These forms do not give a clear answer on their own and can't be evaluated at face value. Therefore, additional methods like L'Hôpital's Rule, factorization, or algebraic manipulation are often required to resolve them. In the ratio \(\frac{\ln x}{\ln (\ln x)}\) as x approaches infinity, we encounter the indeterminate form \(\frac{\infty}{\infty}\), which is why L'Hôpital's Rule is used in the problem to proceed with finding the limit.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a logarithm to the base \text{e}\, where e is an irrational and transcendental constant approximately equal to 2.71828. This particular logarithm is widespread in calculus because it has properties that are fundamental to the study of exponential growth, integration, and the behavior of functions as they tend to infinity.

When we evaluate the natural logarithm of a number, what we're effectively determining is the exponent that e must be raised to in order to get that number. This function grows slower than polynomial and exponential functions, which is why in our exercise, the function \(\ln x\) is revealed to grow faster than \(\ln (\ln x)\), emphasizing the multiplicative relationship between rates of growth when dealing with logarithmic functions.