We are tasked with evaluating the following limit:

$\lim_{x \to 0} \left( \cos(x) \right)^{\frac{3}{x^2}}.$

Step-by-step approach:

1. Take the natural logarithm of the expression:
   To simplify this type of limit, we start by taking the natural logarithm of the expression:

   $y = \left( \cos(x) \right)^{\frac{3}{x^2}} \implies \ln(y) = \frac{3}{x^2} \ln(\cos(x)).$

   Now, we need to evaluate the limit of $\ln(y)$ as $x \to 0$.

2. Evaluate the limit of $\ln(\cos(x))$:
   As $x \to 0$, $\cos(x) \to 1$, and $\ln(\cos(x))$ approaches $\ln(1) = 0$. To get a more precise approximation near 0, we use the Taylor series expansion of $\cos(x)$ around $x = 0$:

   $\cos(x) \approx 1 - \frac{x^2}{2} + O(x^4).$

   Using this, we approximate $\ln(\cos(x))$ for small $x$:

   $\ln(\cos(x)) \approx \ln\left(1 - \frac{x^2}{2}\right).$

   For small $x$, we can use the approximation $\ln(1 + z) \approx z$ when $z$ is small. Therefore:

   $\ln(\cos(x)) \approx -\frac{x^2}{2}.$

3. Substitute this into the expression for $\ln(y)$:
   Now substitute this approximation into the expression for $\ln(y)$:

   $\ln(y) = \frac{3}{x^2} \left( -\frac{x^2}{2} \right) = -\frac{3}{2}.$

4. Exponentiate to find $y$:
   To find $y$, we exponentiate both sides:

   $y = e^{-\frac{3}{2}}.$

Thus, the value of the original limit is:

$\lim_{x \to 0} \left( \cos(x) \right)^{\frac{3}{x^2}} = e^{-\frac{3}{2}}.$

Final Answer:

$\boxed{e^{-\frac{3}{2}}}.$

Would you like more details on any step, or do you have any questions about this approach?

Here are some related questions you might find interesting:

1. How can we compute the limit of $\left( \sin(x) \right)^{\frac{3}{x^2}}$ as $x \to 0$?
2. What is the Taylor series expansion of $\cos(x)$ near $x = 0$?
3. How do you handle limits involving indeterminate forms like $0^\infty$?
4. What are some other techniques for solving limits of exponential functions?
5. How does L'Hopital's rule apply to this kind of limit?

Tip: When dealing with limits of the form $f(x)^{g(x)}$, taking the logarithm of both sides often simplifies the process, especially when $f(x)$ approaches 1 and $g(x)$ grows large.