We are tasked with evaluating the limit:

$\lim_{x \to 0} x^6 \cos\left(\frac{8}{x}\right)$

Step-by-Step Analysis:

1. Oscillatory Nature of $\cos \left( \frac{8}{x} \right)$: The function $\cos \left( \frac{8}{x} \right)$ oscillates between -1 and 1 as $x \to 0$ because the cosine function is periodic. Even though $\frac{8}{x}$ grows without bound as $x$ approaches 0, $\cos$ remains bounded by [-1, 1].

2. Behavior of $x^6$ near $x = 0$: The term $x^6$ tends to 0 as $x \to 0$. Specifically, since $x^6$ shrinks to 0 much faster than $\cos \left( \frac{8}{x} \right)$ oscillates, the entire product $x^6 \cos\left(\frac{8}{x}\right)$ should tend to 0.

3. Squeeze Theorem Application: Since $\cos \left( \frac{8}{x} \right)$ is bounded by -1 and 1, we can apply the Squeeze Theorem: $-x^6 \leq x^6 \cos\left(\frac{8}{x}\right) \leq x^6$ As $x^6 \to 0$ when $x \to 0$, both bounds $-x^6$ and $x^6$ tend to 0. By the Squeeze Theorem, the limit of $x^6 \cos\left(\frac{8}{x}\right)$ as $x \to 0$ is also 0.

Conclusion:

$\lim_{x \to 0} x^6 \cos\left(\frac{8}{x}\right) = 0$

Would you like more details on any of the steps? Here are 5 related questions for further exploration:

1. How does the Squeeze Theorem work in determining limits of oscillatory functions?
2. What are other cases where the cosine function causes oscillation, and how are they handled in limits?
3. Can the behavior of $x^n$ as $x \to 0$ always dominate oscillations like this?
4. What would happen to the limit if the exponent of $x$ were smaller than 6 (e.g., $x^2$)?
5. How would the limit change if the function inside the cosine were different, like $\cos\left( \frac{1}{x} \right)$?

Tip: In many limit problems with oscillating functions, bounding the oscillatory part can help simplify the problem using the Squeeze Theorem.