To evaluate the limit

$\lim_{x \to 1} \left( 1 + \ln(x) \right)^{(x-1)^{-1}},$

let's follow a structured approach.

Step 1: Recognize the indeterminate form

As $x \to 1$, $\ln(x) \to \ln(1) = 0$. Therefore, the base of the exponentiation, $1 + \ln(x)$, approaches 1. At the same time, the exponent $(x - 1)^{-1}$ tends to infinity as $x \to 1$.

This suggests that we have an indeterminate form of the type $1^\infty$. To handle this type of limit, we can rewrite it in a form that makes it easier to evaluate.

Step 2: Take the natural logarithm

Let $y = \left( 1 + \ln(x) \right)^{(x-1)^{-1}}$. Taking the natural logarithm of both sides gives:

$\ln(y) = (x - 1)^{-1} \ln\left( 1 + \ln(x) \right).$

We now need to evaluate the limit of $\ln(y)$ as $x \to 1$, i.e.,

$\lim_{x \to 1} (x - 1)^{-1} \ln\left( 1 + \ln(x) \right).$

Step 3: Apply L'Hopital's Rule

The expression $\frac{\ln(1 + \ln(x))}{x - 1}$ is of the form $\frac{0}{0}$, so we can apply L'Hopital's Rule. First, let's differentiate the numerator and denominator:

• The derivative of $\ln(1 + \ln(x))$ with respect to $x$ is:

$\frac{d}{dx} \left( \ln(1 + \ln(x)) \right) = \frac{1}{1 + \ln(x)} \cdot \frac{1}{x}.$

• The derivative of $x - 1$ with respect to $x$ is simply 1.

Thus, applying L'Hopital's Rule gives:

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

Step 4: Evaluate the limit

As $x \to 1$, $\ln(x) \to 0$, so the expression simplifies to:

$\frac{1}{1(1 + 0)} = 1.$

Thus,

$\lim_{x \to 1} \ln(y) = 1.$

Step 5: Exponentiate to find $y$

Since $\ln(y) \to 1$, it follows that:

$y \to e^1 = e.$

Thus, the original limit is:

$\lim_{x \to 1} \left( 1 + \ln(x) \right)^{(x-1)^{-1}} = e.$

Final Answer:

$\boxed{e}.$

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Would you like to dive deeper into any of the steps or have any questions about the process? Here are some related questions for further exploration:

1. What happens when we apply L'Hopital's Rule to a limit involving an exponential form like $a^{f(x)}$?
2. How can you handle limits of indeterminate forms like $0^0$, $\infty^0$, or $1^\infty$?
3. What is the general procedure for evaluating limits involving logarithms and powers?
4. How does the natural logarithm help in simplifying complex limits?
5. What are some other real-world applications of limits involving logarithmic and exponential functions?

Tip: When encountering limits of the form $1^\infty$, taking the natural logarithm is a useful technique, as it often simplifies the expression and makes it possible to apply L'Hopital’s Rule.