Let's first address the limit problem presented in the image:

The expression is:

$\lim_{{x \to 0}} \left( \cos(2x) \right)^{\frac{1}{x^2}}$

To solve this limit, we need to analyze the behavior of the expression as $x$ approaches 0.

Step 1: Simplify the Cosine Expression

As $x$ approaches 0, $\cos(2x)$ approaches $\cos(0) = 1$. However, because the exponent is $\frac{1}{x^2}$, we need to be more careful.

Step 2: Take the Natural Logarithm

To evaluate the limit, we can take the natural logarithm:

$L = \lim_{{x \to 0}} \ln \left( \left(\cos(2x)\right)^{\frac{1}{x^2}} \right) = \lim_{{x \to 0}} \frac{1}{x^2} \ln \left( \cos(2x) \right)$

Step 3: Expand $\ln(\cos(2x))$ using Taylor Series

Near $x = 0$, $\cos(2x)$ can be approximated by its Taylor series:

$\cos(2x) \approx 1 - 2x^2$

So,

$\ln(\cos(2x)) \approx \ln\left(1 - 2x^2\right)$

Using the approximation $\ln(1 + y) \approx y$ when $y$ is small,

$\ln\left(1 - 2x^2\right) \approx -2x^2$

Thus,

$L = \lim_{{x \to 0}} \frac{1}{x^2} \cdot (-2x^2) = \lim_{{x \to 0}} -2 = -2$

Step 4: Exponentiate to Find the Original Limit

The original limit can now be found by exponentiating:

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

So the limit is:

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

Summary

The limit $\lim_{{x \to 0}} \left( \cos(2x) \right)^{\frac{1}{x^2}}$ evaluates to $e^{-2}$.

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Would you like details on the other questions presented in the image? If so, please let me know!

Here are 8 related questions for further exploration:

1. How does the Taylor series expansion of trigonometric functions help in solving limits?
2. What is the general form of L'Hopital's Rule, and when is it applied?
3. What are common strategies for evaluating limits that approach indeterminate forms?
4. How can you determine if a function is continuous at a given point?
5. What is the significance of the exponential function in calculus, particularly $e^x$?
6. How do you solve limits involving logarithmic and exponential functions?
7. How does the approximation $\ln(1 + y) \approx y$ work, and under what conditions is it valid?
8. Can the limit above be evaluated using series expansion directly without logarithms?

Tip: When dealing with limits involving exponents, taking the natural logarithm can often simplify the problem by transforming the exponential limit into a product or sum, which is easier to evaluate.