Problem 54 - Thomas Calculus in SI Units

Question

Find the limits $$\lim _{x \rightarrow e^{+}}(\ln x)^{1 /(x-e)}$$

Step-by-Step Solution

Answer

The limit is \( \exp\left(\frac{1}{e}\right) \).

Step 1: Recognize Indeterminate Form

The exercise asks us to find \( \lim _{x \rightarrow e^{+}}(\ln x)^{1 /(x-e)} \). Observing this expression, as \( x \to e^+ \), \( \ln(x) \to \ln(e) = 1 \), which means the base approaches 1. The exponent \( \frac{1}{x-e} \) approaches infinity as \( x \to e^+ \), leading to an indeterminate form \( 1^{+\infty} \). We will need another approach to evaluate it.

Step 2: Use Logarithms to Simplify

To resolve the indeterminate form, we take the natural logarithm of the expression:\[\ln\left( (\ln x)^{1/(x-e)} \right) = \frac{1}{x-e} \cdot \ln(\ln x)\]This step helps in transforming the indeterminate form so that we can analyze the behavior of the logarithm.

Step 3: Change of Variables

To handle the limit, perform a change of variables: let \( u = x - e \), then as \( x \to e^+ \), \( u \to 0^+ \). Therefore, our expression transforms:\[\frac{1}{u} \cdot \ln(\ln(e + u))\]

Step 4: Expand with Series

Expand \( \ln(e + u) \) using a Taylor series expansion:\[\ln(e + u) = \ln(e) + \frac{u}{e} + O(u^2) \]Thus, \( \ln(\ln(e+u)) = \ln\left(1 + \frac{u}{e}\right) \approx \frac{u}{e} \).

Step 5: Substitute and Simplify

Substitute the approximation back into the expression from Step 3:\[\frac{1}{u} \cdot \frac{u}{e} = \frac{1}{e}\]As \( u \to 0^+ \), this expression converges to \( \frac{1}{e} \).

Step 6: Exponentiate the Result

Remember that we initially took the natural logarithm of the limit expression. Now, to find the limit itself, we need to exponentiate the result from Step 5:\[\exp\left(\frac{1}{e}\right)\]

Step 7: Conclusion

The result of the original limit is \( \exp\left(\frac{1}{e}\right) \). No further simplification is possible for this limit.

Key Concepts

Indeterminate Forms

When solving limits, you may encounter expressions that do not directly lead to a clear result, known as indeterminate forms. In such cases, the form of the expression does not immediately suggest a specific value. Particularly in this exercise, the expression \((\ln x)^{1 /(x-e)}\) as \(x\) approaches \(e^+\) resolves to the indeterminate form \(1^{+\infty}\). This occurs because:

• The base, \(\ln x\), tends towards 1 as \(x\) approaches \(e\) from the right.
• The exponent, \(\frac{1}{x-e}\), grows towards infinity since \(x-e\) approaches zero from the positive side.

Recognizing an indeterminate form signals that a different method or transformation is needed to evaluate the limit correctly.
Taking the natural logarithm or using L'Hopital's Rule are among the common techniques to resolve indeterminate forms.

Natural Logarithms

The natural logarithm, denoted as \(\ln\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It is a critical function in calculus, particularly when dealing with exponential growth processes or decay phenomena. In this limit exercise, we take advantage of natural logarithms to simplify expressions:

• Natural logarithms convert products into sums, aiding in the application of calculus rules.
• They turn powers into products, which can significantly simplify the resolution of indeterminate forms.

In the solution, we utilize the natural logarithm to transform \((\ln x)^{1/(x-e)}\) into \(\frac{1}{x-e} \cdot \ln(\ln x)\), helping us apply further simplifications through calculus techniques such as series expansion.

Taylor Series Expansion

Taylor series expansion is an invaluable tool in calculus for approximating functions that are otherwise difficult to analyze directly. By expressing a function as an infinite sum of terms calculated from the function’s derivatives at a single point, we can approximate the function for values near that point. In this exercise, we apply Taylor series expansion to the function \(\ln(e + u)\). This aids in simplifying the limit calculation:

• For small \(u\), the Taylor expansion of \(\ln(e + u)\) is approximated as \(\ln(e) + \frac{u}{e} + O(u^2)\).
• This simplifies \(\ln(\ln(e + u))\) to approximately \(\frac{u}{e}\), breaking down the complexity of the expression.

Using the approximate expression \(\frac{u}{e}\), we further simplify the entire limit resolution to \(\frac{1}{e}\), allowing for the exponentiation process which eventually leads us to the solution of the entire problem.