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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.5.57
Chapter 7, Problem 7.5.57

Indeterminate Powers and Products
Find the limits in Exercises 53–68.
57. lim (x → 0⁺) x^(-1/ln x)

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1
Recognize that the limit involves an indeterminate form of the type \(x^{f(x)}\) as \(x \to 0^+\). To handle this, rewrite the expression using the exponential and logarithm functions: \(x^{-1/\ln x} = e^{\ln\left(x^{-1/\ln x}\right)}\).
Apply the logarithm power rule inside the exponent: \(\ln\left(x^{-1/\ln x}\right) = -\frac{1}{\ln x} \cdot \ln x\).
Simplify the expression inside the exponent: \(-\frac{1}{\ln x} \cdot \ln x = -1\).
Rewrite the original limit as \(\lim_{x \to 0^+} e^{-1}\), since the exponent simplifies to a constant.
Since \(e^{-1}\) is a constant, the limit is simply \(e^{-1}\). Thus, the limit exists and equals this constant value.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limits Involving Indeterminate Forms

Limits that result in expressions like 0^0, ∞^0, or 1^∞ are called indeterminate forms. These require special techniques such as rewriting the expression or applying logarithms to evaluate the limit accurately.
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Logarithmic Transformation for Limits

Taking the natural logarithm of a function can simplify complex limit expressions, especially those involving exponents. By converting powers into products, it allows the use of limit laws and L'Hôpital's Rule more effectively.
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L'Hôpital's Rule

L'Hôpital's Rule helps evaluate limits that yield indeterminate forms like 0/0 or ∞/∞ by differentiating the numerator and denominator separately. It is often used after logarithmic transformation to find limits of complicated expressions.
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