Question

In Exercises 2.4.2-2.4.40, find the indicated limits. $$ \lim _{x \rightarrow 1+}(\log x)^{\alpha} \log (\log x) $$

Short answer

Question: Determine the limit of the function $(\log x)^{\alpha} \log (\log x)$ as $x$ approaches $1$ from the right side. Answer: The limit of the function $(\log x)^{\alpha} \log (\log x)$ as $x$ approaches $1$ from the right side is $0$.

Step-by-step solution

Identify the form of the limit

Since the problem involves logarithmic functions, we need to recognize that as \(x\) approaches \(1\), \(\log(x)\) approaches \(0\). Therefore, we have the form \(0^\alpha \cdot \log(0)\) as \(x\) approaches \(1\). This is an indeterminate form, so we will need to use some techniques to find the limit.

Apply L'Hôpital's Rule

L'Hôpital's Rule is a useful technique to find limits of functions, particularly for indeterminate forms. In order to apply L'Hôpital's Rule, we must first rewrite the function in a more suitable form for differentiation so that we have the ratio of two differentiable functions. Let's rewrite the given function as follows: $$ f(x) = \frac{\log(\log x)}{(\log x)^{-\alpha}} $$ Now, as \(x\) approaches \(1\), both the numerator and denominator approach \(0\), so we have an indeterminate form of the type \(\frac{0}{0}\). Thus, we can apply L'Hôpital's Rule, which states that if \(\lim _{x\to a}\frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then $$ \lim _{x\to a}\frac{f(x)}{g(x)}=\lim _{x\to a}\frac{f'(x)}{g'(x)} $$ provided the limit on the right side exists. For our function, we need to find the derivatives of the numerator and denominator.

Find the derivatives of the numerator and denominator

First, we differentiate the numerator with respect to \(x\): $$ \frac{d}{dx}[\log(\log x)] = \frac{d}{dx}[\log u] \cdot \frac{du}{dx}, \text{ where } u = \log x $$ With the chain rule, we have $$ \frac{d}{dx}[\log u] = \frac{1}{u}\cdot \frac{du}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} $$ Next, we differentiate the denominator with respect to \(x\): $$ \frac{d}{dx}[(\log x)^{-\alpha}] = -\alpha (\log x)^{-\alpha - 1}\cdot \frac{1}{x} $$

Apply L'Hôpital's Rule

Now, let's apply L'Hôpital's Rule to find the limit: $$ \lim_{x \rightarrow 1+} \frac{\log(\log x)}{(\log x)^{-\alpha}} = \lim_{x \rightarrow 1+} \frac{\frac{1}{\log x} \cdot \frac{1}{x}}{-\alpha (\log x)^{-\alpha - 1}\cdot \frac{1}{x}} $$ Simplifying the expression, we have $$ \lim_{x \rightarrow 1+} \frac{\frac{1}{\log x}}{-\alpha (\log x)^{-\alpha - 1}} = \lim_{x \rightarrow 1+} \frac{-(\log x)^{\alpha + 1}}{\alpha} $$

Evaluate the limit

Now, we can evaluate the limit as \(x\) approaches \(1\): $$ \lim_{x \rightarrow 1+} \frac{-(\log x)^{\alpha + 1}}{\alpha} = \frac{-(\log 1)^{\alpha + 1}}{\alpha} = \frac{-(0)^{\alpha + 1}}{\alpha} = \boxed{0} $$ Hence, the limit of the function \((\log x)^{\alpha} \log (\log x)\) as \(x\) approaches \(1\) from the right side is \(0\).

Key concepts

L'Hôpital's rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits, especially when dealing with indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The rule is named after the French mathematician Guillaume de l'Hôpital. To apply this rule, your expression must first be transformed if necessary so that it looks like one of these indeterminate forms. Once in this form, you can replace the original limit with a new limit of the derivatives of the numerator and the denominator.

Here's a simple step-by-step what the rule states:

• Identify that you have an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• Take the derivative of the numerator.
• Take the derivative of the denominator.
• Evaluate the limit of the new fraction.

It's important to remember that L'Hôpital's Rule can only be applied if both derivatives exist and the resulting limit is determinable. In practical terms, you will find yourself frequently using this rule when working with limits that involve more complex expressions such as logarithms and exponentials.

logarithmic functions

Logarithmic functions, commonly represented as \(\log(x)\), are the inverse of exponential functions. They are essential in mathematics because they transform multiplicative relationships into additive ones, simplifying complex equations. The base of the logarithm can vary, commonly being 10 (common logarithm) or the natural base \(e\) (natural logarithm, often written as \(\ln(x)\)).

Logarithms have several useful properties:

• Logarithm of a product: \(\log(ab) = \log(a) + \log(b)\)
• Logarithm of a quotient: \(\log(\frac{a}{b}) = \log(a) - \log(b)\)
• Logarithm of a power: \(\log(a^b) = b\log(a)\)

When dealing with continuity and differentiability, the logarithmic function is particularly smooth and its derivative is \(\frac{1}{x}\). As such, the derivative of more complex expressions involving logs can be found through the chain rule.

Logarithms are especially important in contexts where things grow proportionally, such as in the originally given expression \((\log x)^{\alpha} \log(\log x)\), where taking the derivative involves multiple uses of differentiation rules.

indeterminate forms

In calculus, a limit can sometimes lead to an expression that doesn’t initially provide useful information about the behavior of the function. These expressions are called indeterminate forms, and they require additional calculations to determine the actual limit. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \times \infty\), \(\infty - \infty\), \(0^0\), \(1^\infty\), and \(\infty^0\).

Finding the limit involves steps like rewriting expressions, factoring, or using L'Hôpital's Rule. With L'Hôpital's Rule, you specifically need the indeterminate forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) to apply it directly.

In the given problem, \((\log x)^\alpha \log(\log x)\) as \(x\) approaches 1 yields the form \(0^\alpha \cdot (-\infty)\), which resembles \(0 \times \infty\), a complex type of indeterminate form. This demands reformulating to a ratio that can be differentiated using L'Hôpital's Rule.

• This involves clever manipulation such as converting the product into the form \(\frac{0}{0}\) by modifying expressions using algebraic operations.
• The focus is on creating a structure suitable for L'Hôpital's Rule, leading us to eventually use derivatives to resolve the limit.

Understanding these forms and techniques allows for the resolution of limits that initially seem unsolvable.