We need to evaluate the given limit:

$\lim_{x \to 0^+} x^{\frac{\ln 2}{1 + \ln x}}$

Step 1: Simplify the exponent

The term $x^{\frac{\ln 2}{1 + \ln x}}$ can be rewritten using the property $a^{b} = e^{b \ln a}$: $x^{\frac{\ln 2}{1 + \ln x}} = e^{\frac{\ln 2}{1 + \ln x} \ln x}.$

So, the limit becomes: $\lim_{x \to 0^+} e^{\frac{\ln 2 \cdot \ln x}{1 + \ln x}}.$

Step 2: Analyze the exponent

Now focus on the exponent: $\frac{\ln 2 \cdot \ln x}{1 + \ln x}.$

As $x \to 0^+$, $\ln x \to -\infty$. Thus, analyze the denominator $1 + \ln x$, which also tends to $-\infty$. The behavior of the fraction needs closer inspection.

Simplify the fraction:

\frac{\ln 2 \cdot \ln x}{1 + \ln x} \approx \ln 2 \cdot \frac{\ln x}{\ln x} = \ln 2 \quad \text{(as x \to 0^+)}.

Step 3: Evaluate the limit

Thus, the exponent converges to a finite value: $\frac{\ln 2 \cdot \ln x}{1 + \ln x} \to \ln 2.$

Therefore, the limit becomes: $e^{\ln 2} = 2.$

Final Answer:

$\lim_{x \to 0^+} x^{\frac{\ln 2}{1 + \ln x}} = 2.$

Would you like a detailed breakdown of any step or further examples?

Here are 5 follow-up questions to deepen understanding:

1. How does the property $a^b = e^{b \ln a}$ help in limit problems like this?
2. Why does $\ln x \to -\infty$ as $x \to 0^+$?
3. What happens if the numerator of the exponent is more complex, such as $\ln x \cdot \ln x$?
4. How would the solution change if $\ln 2$ were replaced with another constant, like $\ln 3$?
5. What role does L’Hopital’s Rule play in problems involving limits of forms $\frac{-\infty}{\infty}$?

Tip: Always rewrite exponential expressions into their logarithmic form to simplify limits involving powers!