We aim to compute the limit:

$\lim_{x \to 0} \frac{1}{x} \int_0^x \left(1 - \tan(2t)\right)^{\frac{1}{t}} \, dt.$

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Step 1: Analyze the expression inside the integral

The term inside the integral is $\left(1 - \tan(2t)\right)^{\frac{1}{t}}$. As $t \to 0$:

• $\tan(2t) \to 2t$ (using the small-angle approximation for $\tan(x)$).
• Thus, $1 - \tan(2t) \to 1 - 2t$.

Now, approximate $\left(1 - 2t\right)^{\frac{1}{t}}$ using logarithmic expansion: $\ln\left(\left(1 - 2t\right)^{\frac{1}{t}}\right) = \frac{1}{t} \ln(1 - 2t).$ Expand $\ln(1 - 2t)$ for small $t$: $\ln(1 - 2t) \approx -2t.$ So: $\frac{1}{t} \ln(1 - 2t) \approx -2.$ Exponentiating: $\left(1 - \tan(2t)\right)^{\frac{1}{t}} \approx e^{-2}.$

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Step 2: Approximate the integral

Substitute the approximation $\left(1 - \tan(2t)\right)^{\frac{1}{t}} \approx e^{-2}$ into the integral: $\int_0^x \left(1 - \tan(2t)\right)^{\frac{1}{t}} \, dt \approx \int_0^x e^{-2} \, dt.$ Since $e^{-2}$ is constant, this becomes: $\int_0^x e^{-2} \, dt = e^{-2} \int_0^x 1 \, dt = e^{-2} \cdot x.$

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Step 3: Evaluate the limit

Now substitute the result of the integral into the limit: $\lim_{x \to 0} \frac{1}{x} \int_0^x \left(1 - \tan(2t)\right)^{\frac{1}{t}} \, dt \approx \lim_{x \to 0} \frac{1}{x} \cdot e^{-2} x.$ Simplify: $\lim_{x \to 0} e^{-2} = e^{-2}.$

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Final Answer:

$\boxed{e^{-2}}$

Would you like a more detailed breakdown or additional questions related to this problem? Here are some related questions:

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1. How is the small-angle approximation for $\tan(x)$ derived?
2. What is the Taylor expansion of $\ln(1 - x)$, and how is it used here?
3. How do we rigorously justify the substitution of an approximate function into a definite integral?
4. Could this problem be solved using numerical methods or software tools? If so, how?
5. How does this limit relate to exponential growth or decay in applied contexts?

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Tip: When solving integrals with limits, carefully analyze the behavior of the integrand near the limits of integration to identify dominant terms or patterns.