To evaluate the limit:

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

Let's break this down step by step.

Step 1: Behavior of $\sin\left(\frac{1}{x - 8}\right)$

As $x \to \infty$, $\frac{1}{x - 8}$ approaches $0$, and since $\sin(z) \approx z$ for small $z$, we have:

$\sin\left(\frac{1}{x - 8}\right) \approx \frac{1}{x - 8} \quad \text{as} \quad x \to \infty$

So, for very large $x$, we can approximate:

$\sin\left(\frac{1}{x - 8}\right) \approx \frac{1}{x}$

Step 2: Behavior of $(x + 2) \sin\left(\frac{1}{x - 8}\right)$

Now, substituting the approximation $\sin\left(\frac{1}{x - 8}\right) \approx \frac{1}{x}$, we get:

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

This simplifies to:

$(x + 2) \cdot \frac{1}{x} = 1 + \frac{2}{x}$

As $x \to \infty$, $\frac{2}{x} \to 0$, so the expression tends to:

$1 + 0 = 1$

Conclusion:

The limit exists and is equal to 1.

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

Do you want further details or have any questions?

Here are 5 related questions:

1. How does the small angle approximation of sine work for limits?
2. What happens if the sine function argument does not approach zero?
3. How do we approach limits that involve both trigonometric functions and polynomial growth?
4. What are the techniques to simplify limits involving infinity?
5. Can we solve this limit without using approximations?

Tip: When handling limits involving trigonometric functions, approximations like $\sin(x) \approx x$ for small $x$ can simplify the process.