We are tasked with evaluating the following limit:

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

Step-by-step solution:

1. Examine the behavior of the cosine function: $\cos(7x)$ The cosine function is continuous everywhere, and specifically, at $x = 0$, we have: $\cos(7 \cdot 0) = \cos(0) = 1$

2. Simplifying the expression: We now rewrite the limit, knowing that $\cos(7x)$ approaches 1 as $x \to 0$: $\lim_{x \to 0} x^6 \cos(7x) = \lim_{x \to 0} x^6 \cdot 1 = \lim_{x \to 0} x^6$

3. Evaluate the limit of $x^6$ as $x \to 0$: The function $x^6$ approaches 0 as $x$ approaches 0: $\lim_{x \to 0} x^6 = 0$

Therefore, the limit is:

$\boxed{0}$

Would you like any more details or clarifications?

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Here are 5 related questions you might find interesting:

1. How do you find the limit of a product involving trigonometric functions?
2. What is the limit of $x^2 \sin(x)$ as $x \to 0$?
3. How does L'Hôpital's Rule apply to limits involving indeterminate forms?
4. What happens when you take the limit of higher powers of $x$, like $x^8$, as $x \to 0$?
5. Can you explain how continuity helps in simplifying limits?

Tip: When dealing with limits, always check if the trigonometric functions can be simplified near 0 by using known values or approximations (like $\cos(0) = 1$ or $\sin(0) = 0$).