We are tasked with evaluating the following limit:

$\lim_{x \to 0^+} \left(\cos(4x)\right)^{\frac{1}{x^2}}.$

Step-by-step Solution:

1. Rewrite the expression using the logarithm: To simplify the computation, we take the natural logarithm: $y = \left(\cos(4x)\right)^{\frac{1}{x^2}} \implies \ln y = \frac{1}{x^2} \ln \cos(4x).$

   The limit now becomes: $\ln y = \lim_{x \to 0^+} \frac{\ln \cos(4x)}{x^2}.$

2. Expand $\ln \cos(4x)$ using a Taylor series: For small $x$, $\cos(4x) \approx 1 - \frac{(4x)^2}{2} = 1 - 8x^2$. Thus: $\ln \cos(4x) \approx \ln\left(1 - 8x^2\right).$

   Using the approximation $\ln(1 - u) \approx -u$ for small $u$, we have: $\ln \cos(4x) \approx -8x^2.$

3. Substitute back into the limit: Replace $\ln \cos(4x)$ with $-8x^2$ in the limit: $\ln y = \lim_{x \to 0^+} \frac{-8x^2}{x^2}.$

   Simplify: $\ln y = \lim_{x \to 0^+} -8 = -8.$

4. Exponentiate to find $y$: Recall that $y = e^{\ln y}$. Thus: $y = e^{-8}.$

Final Answer:

$\lim_{x \to 0^+} \left(\cos(4x)\right)^{\frac{1}{x^2}} = e^{-8}.$

Let me know if you'd like further clarification or details! Here are 5 related questions to explore:

1. How does the Taylor expansion simplify limits involving trigonometric functions?
2. What is the significance of the natural logarithm in solving exponential limits?
3. How can this method be generalized to evaluate similar limits with $\sin(x)$ or $\tan(x)$?
4. Why does $\ln(1 - u) \approx -u$ hold for small $u$?
5. What are the key conditions for applying this logarithmic transformation to a limit?

Tip: Always verify the convergence of the function inside the logarithm when using logarithmic transformations