To solve the given limit using the second remarkable limit (2nd fundamental limit), let’s analyze the expression step by step.

The given limit is:

$\lim_{x \to 0} \left( \ln(e x + e) \right)^{1 / \sqrt[4]{x}}.$

Step 1: Simplify $\ln(e x + e)$

Factor $e$ out inside the logarithm:

$\ln(e x + e) = \ln(e (x + 1)) = \ln(e) + \ln(x + 1).$

Since $\ln(e) = 1$, we have:

$\ln(e x + e) = 1 + \ln(x + 1).$

Step 2: Rewrite the limit

Substitute $\ln(e x + e)$ back into the limit:

[ \lim_{x \to 0} \left( \ln(e x + e) \right)^{1 / \sqrt[4]{x}} = \lim_{x \to 0} \left( 1 + \ln(x + 1) \right)^{1 / \sqrt[4]{x}}. ]

Step 3: Expand $\ln(x + 1)$ for small $x$

Using the approximation $\ln(1 + u) \approx u$ for small $u$, we approximate:

$\ln(x + 1) \approx x \quad \text{as} \quad x \to 0.$

Thus:

$\ln(e x + e) \approx 1 + x.$

Step 4: Express in exponential form

The base of the exponent is approximately $1 + x$. Using the exponential approximation $(1 + u)^v \approx e^{u \cdot v}$ for small $u$, rewrite the expression:

[ \lim_{x \to 0} \left(1 + x\right)^{1 / \sqrt[4]{x}} \approx \lim_{x \to 0} e^{x \cdot (1 / \sqrt[4]{x})}. ]

Step 5: Simplify the exponent

The exponent becomes:

$x \cdot \frac{1}{\sqrt[4]{x}} = x^{1 - 1/4} = x^{3/4}.$

Thus the limit becomes:

$\lim_{x \to 0} e^{x^{3/4}}.$

Step 6: Evaluate the limit

As $x \to 0$, $x^{3/4} \to 0$. Therefore, the exponential simplifies:

$e^{x^{3/4}} \to e^0 = 1.$

Final Answer:

$\boxed{1.}$

Would you like additional clarification or to see alternative solution methods?

Related Questions:

1. How can you expand $\ln(1 + x)$ for higher orders of approximation?
2. What is the 2nd remarkable limit and how is it applied in general?
3. How do we approximate $(1 + u)^v$ for small $u$?
4. What would the result be if the limit approached infinity instead of zero?
5. How does the order of magnitude for $x^{3/4}$ affect the convergence?

Tip:

When handling limits with exponents, always try converting to exponential form $e^{f(x)}$ for easier analysis of growth rates.