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

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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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주요 개념

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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.

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.

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.

Indeterminate Powers and ProductsFind the limits in Exercises 53–68.57. lim (x → 0⁺) x^(-1/ln x)