The given mathematical expression is:

$\lim_{x \to 8} (x + 2) \sin\left(\frac{1}{x - 8}\right)$

Step-by-step solution:

1. Observe the behavior near $x = 8$:

   • As $x$ approaches 8, the term $\frac{1}{x - 8}$ becomes very large, leading to oscillations in $\sin\left(\frac{1}{x - 8}\right)$ between -1 and 1.
   • The sine function, $\sin(\theta)$, is bounded between -1 and 1 for all real numbers $\theta$, so $\sin\left(\frac{1}{x - 8}\right)$ also oscillates between -1 and 1.

2. Behavior of the term $(x + 2)$:

   • As $x \to 8$, $(x + 2)$ approaches $8 + 2 = 10$.

3. Use the Squeeze Theorem:

   • Since $\sin\left(\frac{1}{x - 8}\right)$ is bounded between -1 and 1, we can bound the entire expression: $- (x + 2) \leq (x + 2) \sin\left(\frac{1}{x - 8}\right) \leq (x + 2)$
   • As $x \to 8$, both the lower bound and the upper bound approach 10. Therefore, by the Squeeze Theorem: $\lim_{x \to 8} (x + 2) \sin\left(\frac{1}{x - 8}\right) = 10$

Final Answer:

$\lim_{x \to 8} (x + 2) \sin\left(\frac{1}{x - 8}\right) = 10$

Would you like more details on the Squeeze Theorem or how limits with oscillating functions behave?

Here are five related questions to explore this further:

1. How does the Squeeze Theorem work for bounded trigonometric functions?
2. What happens to $\sin\left(\frac{1}{x - a}\right)$ as $x$ approaches $a$?
3. Can limits involving oscillatory functions always be solved using the Squeeze Theorem?
4. How do limits of trigonometric functions behave when combined with linear functions?
5. What are other techniques to solve oscillating limits?

Tip: The Squeeze Theorem is particularly useful when a function is bounded but oscillates infinitely as it approaches a particular point.