Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find each limit. (a) \(\lim _{x \rightarrow 0^{+}} x^{x}\) (b) \(\lim _{x \rightarrow 0^{+}}\left(x^{x}\right)^{x}\) (c) \(\lim _{x \rightarrow 0^{+}} x^{\left(x^{x}\right)}\) (d) \(\lim _{x \rightarrow 0^{+}}\left(\left(x^{x}\right)^{x}\right)^{x}\) (e) \(\lim _{x \rightarrow 0^{+}} x^{\left(x^{\left(x^{r}\right)}\right)}\)

Step-by-Step Solution

Answer

(a) 1, (b) 1, (c) 0, (d) 1, (e) 1 when \( r > 0 \).

Step 1: Simplify the Expression for (a)

To find \( \lim_{x \to 0^{+}} x^{x} \), rewrite the expression using the natural logarithm: \( y = x^{x} = e^{x \ln x} \). Evaluate the limit of the exponent: \( \lim_{x \to 0^{+}} x \ln x \). As \( x \to 0^{+} \), \( \ln x \to -\infty \), thus \( x \ln x \to 0 \). Therefore, \( \lim_{x \to 0^{+}} e^{x \ln x} = e^{0} = 1 \).

Step 2: Simplify the Expression for (b)

Given \( \lim_{x \to 0^{+}} \left(x^{x}\right)^{x} = \lim_{x \to 0^{+}} x^{x^2} \), express the exponent as \( y = x^{x^2} = e^{x^2 \ln x} \). Evaluate \( \lim_{x \to 0^{+}} x^2 \ln x \): as \( x \to 0^{+} \), \( x^2 \ln x \to 0 \). This leads to \( \lim_{x \to 0^{+}} e^{x^2 \ln x} = e^{0} = 1 \).

Step 3: Simplify the Expression for (c)

For \( \lim_{x \to 0^{+}} x^{\left(x^{x}\right)} = x^{1} \), note that this simplifies to \( \lim_{x \to 0^{+}} x = 0 \) since the exponent \( x^x \to 1 \) as \( x \to 0^{+} \).

Step 4: Simplify the Expression for (d)

For \( \lim_{x \to 0^{+}} \left(\left(x^{x}\right)^{x}\right)^{x} = \lim_{x \to 0^{+}} x^{x^{3}} \), express it as \( y = x^{x^{3}} = e^{x^{3} \ln x} \). Calculate \( \lim_{x \to 0^{+}} x^{3} \ln x \): since \( x^{3} \ln x \to 0 \), \( \lim_{x \to 0^{+}} e^{x^{3} \ln x} = e^{0} = 1 \).

Step 5: Simplify the Expression for (e)

Consider \( \lim_{x \to 0^{+}} x^{\left(x^{\left(x^{r}\right)}\right)} \). If \( r > 0 \), \( x^{r} \to 0 \), making \( x^{x^{x^{r}}} \to 1 \) due to the exponent approaching 0. Thus, \( \lim_{x \to 0^{+}} x^{x^{x^{r}}} = 1 \). However, if \( r = 0 \), the configuration fails, but the general calculation \( x \to 1 \) persists. Therefore, for \( r > 0 \), the limit is \( 1 \).

Key Concepts

Exponential Functions

Exponential functions are a type of function where a constant base is raised to a variable exponent. They have a general form of the type \( f(x) = a^{x} \), where \( a \) is a constant.
The behavior of exponential functions can be quite intriguing, as they can grow rapidly or decay swiftly depending on the value of \( a \).

• If \( a > 1 \), the function \( f(x) = a^{x} \) is an increasing function. It grows faster as \( x \) becomes larger.
• If \( 0 < a < 1 \), the function is decreasing, meaning it comes closer to zero as \( x \) increases.

These functions often appear across various fields of study because they model processes involving growth or decay.
For example, populations that grow at a constant percentage per unit time can be modeled with exponential functions.In the given exercise, we observe exponential functions where both the base and the exponent involve the variable \( x \). This complex form requires careful consideration as \( x \) approaches particular values, such as zero in this case, to determine the limit.
It's important to simplify these expressions often using logarithmic transformations to make understanding easier.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base of Euler's number \( e \), which is approximately equal to 2.71828.
The natural logarithm gives us the time needed to grow to a certain level when dealing with continuous growth processes.

• Natural logarithms convert multiplication into addition, which simplifies many algebraic expressions.
• They are particularly useful with exponential functions, as they allow us to "bring down" exponents, making complex functions easier to analyze.

In the step-by-step solution provided, we use the property of the natural logarithm to transform the function \( x^x \) into an exponential form.
This transformation involves expressing \( x^x \) as \( e^{x \ln x} \). This change of perspective makes it easier to analyze how the function behaves as \( x \) approaches zero.This logarithmic transformation reveals that the exponent \( x \ln x \) approaches zero as \( x \) goes to zero from the positive side.
Thus, expressing \( x^x \) in exponential form allows for a much simpler evaluation of its limit.

Limits of Functions

Limits are fundamental concepts in calculus. They help us understand how functions behave as their inputs approach specific values.

• To find the limit of a function is to find its value as the variable approaches a particular point.
• Limits are essential for defining derivatives and integrals, which are core elements of calculus.

In the exercise provided, the focus is on limits of exponential-type functions as \( x \to 0^{+} \). The goal is to determine what these functions approach when \( x \) is nearly zero.
Understanding these limits involves several steps, including applying logarithms to simplify the exponent, as seen in the solutions outlined.The solutions highlight different methods to resolve the limits:

• Transforming the expression using the natural logarithm to linearize the exponent.
• Following through with evaluating these transformed expressions as \( x \) approaches zero.

By identifying whether the resultant expression approaches zero, one, or another finite limit, we gain deeper insights into the behavior of the original complex expression.
These techniques are essential for dealing with indeterminate forms, such as \( 0^0 \), which often arise in these types of exercises.