Let $a_n\ (n\geq 1)$ be the value for which $\int_x^{2x} e^{-t^n}dt\ (x\geq 0)$ is maximal. Find $\lim_{n\to\infty} \ln a_n.$
1. Define the integral \( I_n(x) = \int_x^{2x} e^{-t^n} \, dt \). We need to find the value of \( x \) that maximizes \( I_n(x) \).

2. To find the maximum, we first compute the derivative of \( I_n(x) \) with respect to \( x \):
   \[
   I_n'(x) = \frac{d}{dx} \left( \int_x^{2x} e^{-t^n} \, dt \right)
   \]
   Using the Leibniz rule for differentiation under the integral sign, we get:
   \[
   I_n'(x) = e^{-(2x)^n} \cdot 2 - e^{-x^n}
   \]
   Simplifying, we have:
   \[
   I_n'(x) = 2e^{-(2x)^n} - e^{-x^n}
   \]

3. Set the derivative equal to zero to find the critical points:
   \[
   2e^{-(2x)^n} - e^{-x^n} = 0
   \]
   \[
   2e^{-(2x)^n} = e^{-x^n}
   \]
   Taking the natural logarithm on both sides:
   \[
   \ln(2e^{-(2x)^n}) = \ln(e^{-x^n})
   \]
   \[
   \ln 2 - (2x)^n = -x^n
   \]
   \[
   \ln 2 = (2^n - 1)x^n
   \]
   Solving for \( x^n \):
   \[
   x^n = \frac{\ln 2}{2^n - 1}
   \]
   Therefore:
   \[
   x = \left( \frac{\ln 2}{2^n - 1} \right)^{\frac{1}{n}} = a_n
   \]

4. We need to find \( \lim_{n \to \infty} \ln a_n \):
   \[
   \ln a_n = \ln \left( \left( \frac{\ln 2}{2^n - 1} \right)^{\frac{1}{n}} \right)
   \]
   \[
   \ln a_n = \frac{1}{n} \ln \left( \frac{\ln 2}{2^n - 1} \right)
   \]
   \[
   \ln a_n = \frac{1}{n} \left( \ln (\ln 2) - \ln (2^n - 1) \right)
   \]

5. Evaluate the limit:
   \[
   \lim_{n \to \infty} \ln a_n = \lim_{n \to \infty} \frac{\ln (\ln 2) - \ln (2^n - 1)}{n}
   \]
   Since \( \ln (\ln 2) \) is a constant, we focus on the term involving \( 2^n \):
   \[
   \lim_{n \to \infty} \frac{\ln (\ln 2) - \ln (2^n - 1)}{n} = \lim_{n \to \infty} \frac{-\ln (2^n - 1)}{n}
   \]
   Using the approximation \( \ln (2^n - 1) \approx \ln (2^n) = n \ln 2 \) for large \( n \):
   \[
   \lim_{n \to \infty} \frac{-n \ln 2}{n} = -\ln 2
   \]

The final answer is \(\boxed{-\ln 2}\).