The task is to calculate the limit:

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

Step-by-step solution:

1. Rewrite the expression using logarithms:

To simplify, let's take the natural logarithm: $y = (\cos x)^{\frac{3}{x^2}} \implies \ln y = \frac{3}{x^2} \ln(\cos x).$

The problem now reduces to finding: $\lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{3 \ln(\cos x)}{x^2}.$

2. Expand $\ln(\cos x)$ using Taylor series:

For small $x$, $\cos x \approx 1 - \frac{x^2}{2}$. Taking the natural logarithm: \ln(\cos x) \approx \ln\left(1 - \frac{x^2}{2}\right) \approx -\frac{x^2}{2} \quad \text{(since \ln(1 - u) \approx -u$for small$u)}.

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

3. Substitute back into the limit:

Substituting $\ln(\cos x) \approx -\frac{x^2}{2}$: $\lim_{x \to 0} \frac{3 \ln(\cos x)}{x^2} \approx \lim_{x \to 0} \frac{3 \cdot \left(-\frac{x^2}{2}\right)}{x^2} = \lim_{x \to 0} \frac{-3}{2} = -\frac{3}{2}.$

4. Exponentiate to find $y$:

Since $\ln y = -\frac{3}{2}$, we exponentiate both sides to get: $y = e^{-\frac{3}{2}}.$

Final Answer:

The limit is: $\boxed{e^{-\frac{3}{2}}}.$

Would you like further clarification or details?

Related Questions:

1. How does the Taylor expansion help in approximating $\cos x$ near zero?
2. What are the steps to solve a limit of the form $a^{f(x)}$?
3. Why is the natural logarithm used in this problem?
4. How can L’Hôpital’s rule be applied to logarithmic limits?
5. What happens if we replace $\cos x$ with $\sin x$ in this problem?

Tip:

For problems involving powers of trigonometric functions with small angles, consider logarithmic transformations and series expansions for simplification!