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) \(\lim _{x \rightarrow 0^{+}} x^{\left(x^{\left(x^{*} \mid\right.}\right)}\)

Step-by-Step Solution

Answer

All limits equal 1.

Step 1: Understanding Limit (a)

We need to evaluate \( \lim_{x \to 0^+} x^x \). When \( x \) is very small and positive, \( x^x \) can be expressed as \( e^{x \ln(x)} \). We'll find \( \lim_{x \to 0^+} x \ln(x) \).

Step 2: Evaluating Logarithmic Limit

Consider \( \ln(x) \to -\infty \) as \( x \to 0^+ \). Therefore, \( x \ln(x) \approx x(-\infty) \to 0 \). Hence, \( \lim_{x \to 0^+} x \ln(x) = 0 \).

Step 3: Computing Limit (a)

From Step 2, \( \lim_{x \to 0^+} e^{x\ln(x)} = e^0 = 1 \). So, \( \lim_{x \to 0^+} x^x = 1 \).

Step 4: Understanding Limit (b)

Consider \( \lim_{x \to 0^+} (x^x)^x = x^{x^2} \). We want to evaluate this limit as \( x \to 0^+ \).

Step 5: Evaluating Expression for Limit (b)

Rewrite \( x^{x^2} = e^{x^2 \ln(x)} \). Find \( \lim_{x \to 0^+} x^2 \ln(x) \).

Step 6: Simplifying Logarithmic Expression

As \( x \to 0^+ \), \( \ln(x) \to -\infty \). Thus, \( x^2 \ln(x) \) approaches a small negative number, leading to \( \lim_{x \to 0^+} x^2 \ln(x) = 0 \).

Step 7: Computing Limit (b)

Thus, \( \lim_{x \to 0^+} e^{x^2\ln(x)} = e^0 = 1 \). Therefore, \( \lim_{x \to 0^+} (x^x)^x = 1 \).

Step 8: Understanding Limit (c)

For \( \lim_{x \to 0^+} x^{x^x} \), express \( x^{x^x} = e^{x^x \ln(x)} \). Find \( \lim_{x \to 0^+} x^x \ln(x) \).

Step 9: Simplifying Limit Expression for Limit (c)

As \( x \to 0^+ \), \( x^x \to 1 \) (verified previously), and \( \ln(x) \to -\infty \). Hence, \( x^x \ln(x) \approx 1 \cdot (-\infty) \to 0 \).

Step 10: Computing Limit (c)

This gives \( \lim_{x \to 0^+} e^{x^x \ln(x)} = e^0 = 1 \). Hence, \( \lim_{x \to 0^+} x^{x^x} = 1 \).

Step 11: Understanding Limit (d)

Consider \( \lim_{x \to 0^+} ((x^x)^x)^x = (x^{x^2})^x = x^{x^3} \). We need to evaluate \( \lim_{x \to 0^+} x^{x^3} \).

Step 12: Simplifying Expression for Limit (d)

Re-write \( x^{x^3} = e^{x^3 \ln(x)} \), find \( \lim_{x \to 0^+} x^3 \ln(x) \).

Step 13: Evaluating Limit Expression for Limit (d)

For \( \lim_{x \to 0^+} x^3 \ln(x) \to 0 \), similar to previous steps, because \( x \to 0^+ \) yields \( x^3 \ln(x) \approx 0 \).

Step 14: Computing Limit (d)

Thus, \( \lim_{x \to 0^+} e^{x^3 \ln(x)} = e^0 = 1 \). Therefore, \( \lim_{x \to 0^+} ((x^x)^x)^x = 1 \).

Step 15: Understanding Limit (e)

Consider \( \lim_{x \to 0^+} x^{x^{x^{...}}} \). Assuming the hierarchy converges to a pattern as \( x \to 0^+ \).

Step 16: Evaluating Limit Expression for Limit (e)

The tower \( x^{x^{x^{...}}} \) where the exponents are repeated above a small positive \( x \), can be rewritten in a similar pattern seen in previous expressions and follows the same convergence pattern.

Step 17: Computing Limit (e)

The pattern holds as \( x \to 0^+ \); the exponential terms tend to '0' in the log form and thus \( \lim_{x \to 0^+} x^{x^{x^{...}}} = 1 \).

Key Concepts

Limit of a function

Limits are the foundation of calculus. They describe the behavior of functions as they approach a specific point. When we talk about the "limit of a function," we're interested in what value a function is heading towards as the input gets closer to a certain number. For example, when we say \( \lim_{x \to 0^+} x^x \), it means we want to know what happens to \( x^x \) as \( x \) gets really close to 0 from the positive side.

Understanding limits helps in determining the behavior of functions near points where they might not be explicitly defined. This is especially valuable for functions involving powers and compositions, where straightforward calculations may become undefined or complex. By representing limits mathematically with functions, we simplify expressions and manage their behavior more easily.

In the problems you encounter, substituting problematic points often involves leveraging rules of limits or approximations. Here, for functions like \( e^{x\ln(x)} \), knowing that \( x \ln(x) \to 0 \) as \( x \to 0^+ \) allows us to conclude that \( \lim_{x \to 0^+} e^{x\ln(x)} = e^0 = 1 \). This insight shows the power of limits in simplifying and understanding function behavior.

Exponential functions

Exponential functions are a class of functions that involve constant bases raised to variable exponents. Mathematically, they are expressed as \( f(x) = a^x \), where \( a \) is a positive constant. These functions are crucial in calculus as they exhibit rapid growth or decay, making them useful in many real-world applications—from population models to the decay of radioactive materials.

In the context of the limits we evaluated, exponential forms like \( x^x \) can be rewritten using exponential notation: \( x^x = e^{x\ln(x)} \). This transformation is useful because it allows us to deal with exponents as linear expressions (using the natural logarithm), making it easier to apply limits. The exponential function, specifically \( e^x \), serves as a building block in calculus, often appearing in derivatives and integrals due to its unique property of being equal to its own derivative.

These properties allow us to explore and solve complex limit problems. By converting a problem involving powers or nested exponents into an exponential form, we can manage and compute its behavior efficiently as we did with the various forms of \( x^x \) expressions.

Logarithmic functions

Logarithmic functions are the inverses of exponential functions. The logarithm of a number is the exponent to which a base (commonly \( e \) or 10) must be raised to obtain that number. For example, \( \, \log_e(x) \) (also written as \( \ln(x) \)) is used widely because of its natural properties in calculus.

Logarithmic functions are particularly helpful in converting complex exponentials into linear problems—this is their main use in limit problems, like \( x\ln(x) \). By transforming \( x^x \) into \( e^{x\ln(x)} \), the logarithmic part \( x\ln(x) \) becomes an essential component in understanding how the function behaves as \( x \) approaches certain limits.

As \( x \to 0^+ \), \( \ln(x) \to -\infty \), which can complicate direct evaluations. However, by combining \( \ln(x) \) with a factor that approaches zero like \( x^2 \) or \( x^3 \) in our expressions, the product approaches zero: \( x^2\ln(x) \to 0 \). This simplification helps us understand and calculate limits successfully, showing how logarithms can make an exponential function's behavior near critical points more tangible.