We are tasked with evaluating the following limit:

$\lim_{x \to 0} x^6 \cos\left(\frac{7}{x}\right)$

Step-by-step solution:

1. Behavior of the cosine function:
   The function $\cos\left(\frac{7}{x}\right)$ oscillates between -1 and 1 for all values of $x \neq 0$, since cosine is a bounded function. Therefore, we can write:

   $-1 \leq \cos\left(\frac{7}{x}\right) \leq 1$

2. Multiplying by $x^6$:
   Now, multiply both sides of the inequality by $x^6$:

   $-x^6 \leq x^6 \cos\left(\frac{7}{x}\right) \leq x^6$

3. Taking the limit:
   As $x \to 0$, $x^6 \to 0$. This applies to both the upper and lower bounds of the inequality:

   $\lim_{x \to 0} -x^6 = 0 \quad \text{and} \quad \lim_{x \to 0} x^6 = 0$

   Therefore, by the squeeze theorem:

   $\lim_{x \to 0} x^6 \cos\left(\frac{7}{x}\right) = 0$

Final Answer:

$\boxed{0}$

Would you like more details or have any questions? Here are five related questions for further exploration:

1. How does the squeeze theorem work, and in which situations can it be applied?
2. Can we use the same approach to find limits for functions like $x^n \sin\left(\frac{a}{x}\right)$ as $x \to 0$?
3. Why does cosine oscillate but not affect the limit result in this case?
4. What would happen if the exponent of $x$ was less than 1, say $x^2$, instead of $x^6$?
5. How would this limit change if it involved a function like $\sin\left(\frac{7}{x}\right)$ instead of cosine?

Tip: The squeeze theorem is particularly useful for functions with oscillatory terms like sine or cosine when they are bounded and multiplied by a term approaching zero.