To find the limit:

$\lim_{x \to \infty} \left(1 + 2x\right)^{\frac{1}{2 \ln x}}$

we can proceed with the following steps:

1. Rewrite the Expression in Exponential Form: Rewrite the expression as: $\lim_{x \to \infty} e^{\frac{1}{2 \ln x} \cdot \ln(1 + 2x)}$ So now we need to evaluate: $\lim_{x \to \infty} \frac{\ln(1 + 2x)}{2 \ln x}$

2. Analyze the Fraction Inside the Exponent: Consider the limit: $\lim_{x \to \infty} \frac{\ln(1 + 2x)}{2 \ln x}$ As $x \to \infty$, $\ln(1 + 2x)$ behaves like $\ln(2x)$, which can be rewritten as $\ln 2 + \ln x$. Thus: $\lim_{x \to \infty} \frac{\ln(1 + 2x)}{2 \ln x} = \lim_{x \to \infty} \frac{\ln(2) + \ln(x)}{2 \ln x}$

3. Separate the Terms in the Limit: Split the limit as follows: $= \lim_{x \to \infty} \frac{\ln 2}{2 \ln x} + \lim_{x \to \infty} \frac{\ln x}{2 \ln x}$ The first term, $\frac{\ln 2}{2 \ln x}$, approaches 0 as $x \to \infty$. The second term simplifies to: $\lim_{x \to \infty} \frac{\ln x}{2 \ln x} = \frac{1}{2}$

4. Combine Results: Therefore: $\lim_{x \to \infty} \frac{\ln(1 + 2x)}{2 \ln x} = 0 + \frac{1}{2} = \frac{1}{2}$

5. Apply the Result: Substitute this result back into the exponential expression: $\lim_{x \to \infty} e^{\frac{1}{2 \ln x} \cdot \ln(1 + 2x)} = e^{\frac{1}{2}} = \sqrt{e}$

Final Answer:

$\lim_{x \to \infty} \left(1 + 2x\right)^{\frac{1}{2 \ln x}} = \sqrt{e}$

Would you like further details on any step?

Here are some related questions to explore:

1. What happens if we change the base of the logarithm in this problem?
2. How would this problem change if the exponent were $\frac{1}{3 \ln x}$ instead?
3. What methods can we use to find limits involving powers of functions?
4. How does L'Hôpital's Rule apply to limits with logarithmic expressions?
5. Why does the approximation $\ln(1 + 2x) \approx \ln(2x)$ work for large $x$?

Tip: Converting expressions with exponents into exponential form $e^{f(x)}$ can simplify complex limits, especially when dealing with logarithms.