To evaluate the limit

$\lim_{x \to 0^+} (\ln x)^{\frac{1}{1 - \ln x}},$

we proceed step-by-step.

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Step 1: Simplify the exponent

The expression can be rewritten as: $y = (\ln x)^{\frac{1}{1 - \ln x}}.$

Take the natural logarithm on both sides: $\ln y = \frac{\ln(\ln x)}{1 - \ln x}.$

The limit now becomes: $\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} \frac{\ln(\ln x)}{1 - \ln x}.$

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Step 2: Analyze behavior of terms as $x \to 0^+$

• As $x \to 0^+$, $\ln x \to -\infty$.
• $\ln(\ln x) \to \ln(-\infty) \to \infty$.
• The denominator $1 - \ln x \to \infty$.

Thus, the expression becomes a form $\frac{\infty}{\infty}$, so we apply L'Hôpital's Rule.

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Step 3: Apply L'Hôpital's Rule

Differentiate numerator and denominator:

• Derivative of the numerator: $\ln(\ln x)$ gives: $\frac{d}{dx} [\ln(\ln x)] = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}.$
• Derivative of the denominator: $1 - \ln x$ gives: $\frac{d}{dx} [1 - \ln x] = -\frac{1}{x}.$

Now the limit becomes: $\lim_{x \to 0^+} \frac{\ln(\ln x)}{1 - \ln x} = \lim_{x \to 0^+} \frac{\frac{1}{x \ln x}}{-\frac{1}{x}} = \lim_{x \to 0^+} \frac{-1}{\ln x}.$

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Step 4: Evaluate simplified limit

As $x \to 0^+$, $\ln x \to -\infty$, so: $\frac{-1}{\ln x} \to 0.$

Thus, $\ln y \to 0$.

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Step 5: Exponentiate the result

Since $\ln y \to 0$, we have $y \to e^0 = 1$.

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Final Answer:

$\lim_{x \to 0^+} (\ln x)^{\frac{1}{1 - \ln x}} = 1.$

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Would you like a detailed breakdown of any step?

Here are 5 related questions to deepen understanding:

1. Why does $\ln x \to -\infty$ as $x \to 0^+$?
2. How is L'Hôpital's Rule applied in the context of logarithmic functions?
3. What happens if $\ln x$ approaches positive infinity instead of negative infinity?
4. Can we generalize this method for other limits involving logarithmic exponents?
5. How does $e^0 = 1$ follow from the properties of exponential functions?

Tip: Always check the domain of logarithmic functions to ensure the variable is valid for evaluation (e.g., $\ln x$ is only defined for $x > 0$).