To evaluate $\lim_{x \to 0^+} \left( \cos(2x) \right)^{1/x^2}$, let's break it down step-by-step.

Step 1: Set up the limit with a logarithmic approach

Define the expression inside the limit as: $f(x) = \left( \cos(2x) \right)^{1/x^2}$ Then we want to evaluate: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left( \cos(2x) \right)^{1/x^2}$

Since we have a form of $[\text{base}]^{\text{power}}$, taking the natural logarithm will make it easier to handle the exponent. Let $L = \lim_{x \to 0^+} f(x)$. Then: $\ln L = \lim_{x \to 0^+} \frac{\ln \cos(2x)}{x^2}$

Step 2: Analyze $\cos(2x)$ as $x \to 0^+$

As $x \to 0$, we know that $\cos(2x)$ approaches $\cos(0) = 1$. However, to analyze this limit more closely, we can use the Taylor expansion of $\cos(2x)$ around $x = 0$: $\cos(2x) \approx 1 - \frac{(2x)^2}{2} = 1 - 2x^2$

Thus, for small $x$, we can approximate: $\cos(2x) \approx 1 - 2x^2$

Step 3: Substitute into the logarithmic form

Now, substitute this approximation into the logarithmic expression: $\ln L = \lim_{x \to 0^+} \frac{\ln(1 - 2x^2)}{x^2}$

Step 4: Use the approximation $\ln(1 - u) \approx -u$ for small $u$

For values of $u$ close to zero, $\ln(1 - u) \approx -u$. Here, $u = 2x^2$, so: $\ln(1 - 2x^2) \approx -2x^2$

Substituting this into our expression: $\ln L = \lim_{x \to 0^+} \frac{-2x^2}{x^2}$

Step 5: Simplify and evaluate the limit

Simplify the expression: $\ln L = \lim_{x \to 0^+} -2 = -2$

Thus, $\ln L = -2$, which implies that $L = e^{-2}$.

Conclusion

$\lim_{x \to 0^+} \left( \cos(2x) \right)^{1/x^2} = e^{-2}$

Would you like more details, or do you have any questions?

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1. How would the approach change if the function inside the limit involved a sine function?
2. What if the exponent were $\frac{1}{x}$ instead of $\frac{1}{x^2}$?
3. How can we apply Taylor series to analyze limits more generally?
4. What other approximation techniques might be useful in limits involving trigonometric functions?
5. How does L'Hôpital's rule compare as an alternative method for this problem?

Tip: When dealing with expressions of the form $[\text{base}]^{\text{power}}$ in limits, taking the logarithm can simplify the problem by turning the exponentiation into multiplication.