Problem 58 - Calculus - AP Edition

Question

There is a myth that circulates among beginning calculus students which states that all indeterminate forms of types \(0^{0}, \infty^{0},\) and \(1^{\infty}\) have value 1 because "anything to the zero power is \(1^{\prime \prime}\) and \(^{\prime \prime} 1\) to any power is \(1 . "\) The fallacy is that \(0^{0}, \infty^{0},\) and \(1^{\infty}\) are not powers of numbers, but rather descriptions of limits. The following examples, which were suggested by Prof. Jack Staib of Drexel University, show that such indeterminate forms can have any positive real value: (a) \(\lim _{x \rightarrow 0^{+}}\left[x^{(\ln a) /(1+\ln x)}\right]=a \quad\left(\text { form } 0^{0}\right)\) (b) \(\lim _{x \rightarrow+\infty}\left[x^{(\ln a) /(1+\ln x)}\right]=a \quad\left(\text { form } \infty^{0}\right)\) (c) \(\lim _{x \rightarrow 0}\left[(x+1)^{(\ln a) / x}\right]=a \quad\left(\text { form } 1^{\infty}\right)\). Verify these results.

Step-by-Step Solution

Answer

The limits for all parts (a), (b), and (c) confirm that each indeterminate form can equal a positive real number \( a \).

Step 1: Analyze the Indeterminate Form in Part (a)

For part (a), we need to evaluate the limit \( \lim_{x \rightarrow 0^{+}} \left[ x^{(\ln a) /(1+\ln x)} \right] \). This appears to be of the form \( 0^0 \). To analyze this further, let's set \( f(x) = x^{(\ln a) /(1+\ln x)} \) and examine its behavior as \( x \rightarrow 0^{+} \).

Step 2: Apply Logarithmic Transformation for Part (a)

Taking the natural logarithm of \( f(x) \), we have \( \ln f(x) = \frac{\ln a}{1+\ln x} \ln x \). Substitute \( y = \ln x \), which implies \( x = e^y \) and \( y \rightarrow -\infty \) as \( x \rightarrow 0^{+} \). The limit becomes \( \lim_{y \rightarrow -\infty} \frac{\ln a}{1+y} y = \lim_{y \rightarrow -\infty} \frac{\ln a \, y}{1+y} \).

Step 3: Solve the Transformed Limit for Part (a)

Observe that the expression \( \frac{y}{1+y} \) simplifies to \( 1 \) as \( y \) becomes very negative. Thus, the limit reduces to \( \ln a \), leading to \( f(x) = e^{\ln a} = a \). Therefore, \( \lim_{x \rightarrow 0^{+}} x^{(\ln a) /(1+\ln x)} = a \), verifying the result.

Step 4: Analyze Indeterminate Form in Part (b)

For part (b), evaluate \( \lim_{x \rightarrow +\infty} \left[ x^{(\ln a) /(1+\ln x)} \right] \). This appears as an indeterminate form \( \infty^0 \). Again, set \( f(x) = x^{(\ln a) /(1+\ln x)} \) and find \( \ln f(x) = \frac{\ln a}{1+\ln x} \ln x \).

Step 5: Apply Substitution for Limit in Part (b)

Applying substitution \( y = \ln x \), we have \( x = e^y \) and \( y \rightarrow +\infty \). The limit becomes \( \lim_{y \rightarrow +\infty} \frac{\ln a \, y}{1+y} \).

Step 6: Solve Transformed Limit for Part (b)

The expression \( \frac{y}{1+y} \) again approaches \( 1 \) as \( y \rightarrow +\infty \). Therefore, the limit evaluates to \( \ln a \) resulting in \( f(x) = e^{\ln a} = a \). Thus, \( \lim_{x \rightarrow +\infty} x^{(\ln a) /(1+\ln x)} = a \).

Step 7: Analyze Indeterminate Form in Part (c)

For part (c), evaluate \( \lim_{x \rightarrow 0} \left[ (x+1)^{(\ln a)/x} \right] \), which is the form \( 1^\infty \). Again propose \( f(x) = (x+1)^{(\ln a)/x} \). Taking the natural logarithm gives \( \ln f(x) = \frac{\ln a}{x} \ln(x+1) \).

Step 8: Use L'Hôpital's Rule for Part (c) Limit

Apply the series expansion \( \ln(1+x) \approx x \) for small \( x \), simplifying the limit \( \frac{\ln a}{x} \cdot x = \ln a \). As \( x \rightarrow 0 \), the limit becomes \( \ln a \), leading to \( f(x) = e^{\ln a} = a \).

Step 9: Summarize the Verification for Part (c)

Therefore, \( \lim_{x \rightarrow 0} (x+1)^{(\ln a)/x} = a \), confirming the original statement. All calculations across parts (a), (b), and (c) verify the results as intended.

Key Concepts

Limits in Calculus

Limits in calculus are essential for understanding the behavior of functions as they approach certain points. When we talk about limits, we are interested in what a function is doing as the input - values approach a particular number. Limits can help us pinpoint values that are not easily described by traditional function evaluation means.

A classic use of limits is to evaluate a function at points where the function is undefined or takes on an indeterminate form. For example, when trying to evaluate \( \lim_{x \to 0^{+}} x^{x} \), you cannot simply plug in \( x = 0 \) because it results in the form \( 0^0 \). Instead, you examine the behavior of the function as \( x \) gets closer to \( 0 \) from the positive side.

Thus, understanding limits is about seeing where a function 'spirals' around a particular point, giving us profound insight into the function's behavior that isn't available at face value. This fundamental concept is crucial, especially when dealing with indeterminate forms that can't be solved through direct substitution.

Logarithmic Transformations

Logarithmic transformations are a powerful tool in calculus because they can simplify complex expressions, particularly those involving exponents. By applying a logarithm, we can transform an exponentiation problem into a multiplication problem, making it easier to work with.

For example, in the limits we are examining, taking the natural logarithm of a complex exponential expression allows us to bring the exponent down and handle the expression through more straightforward algebraic means. This is especially helpful when dealing with indeterminate forms where direct evaluation is impossible.

Here is how it works: Suppose you have a function \( f(x) = x^{g(x)} \). By applying a logarithm, you transform this into \( \ln f(x) = g(x) \cdot \ln x \). Once in this form, we can apply algebraic manipulation, substitutions, or L'Hôpital's Rule if needed, to find limits at points where the original function became challenging.

L'Hôpital's Rule

L'Hôpital's Rule is a valuable method for resolving indeterminate forms in calculus. When directly evaluating a limit leads you to forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule can often simplify these expressions considerably.

L'Hôpital's Rule states that if we have a limit of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the limit of the ratio of functions will be the same as the limit of the ratio of their derivatives, provided these limits exist. In more concrete terms and with \( f(x) \) and \( g(x) \) as functions, it is expressed as follow: \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

This means that if direct application of the limit on \( \frac{f(x)}{g(x)} \) does not produce a clear answer, you can differentiate the numerator and the denominator separately, and then try taking the limit again. L'Hôpital's rule can be iteratively applied if needed until the form is determinate.