does the corresponding limit exist? lim (x approaching infinity) (x +2)sin(1/(x-8)

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.

Does the corresponding limit exist? lim (x approaching infinity) (x + 2)sin(1/(x-8))

This page provides a detailed step-by-step solution to finding the limit of (x + 2)sin(1/(x - 8)) as x approaches infinity. It explains key concepts such as the small angle approximation for sine and the behavior of functions at infinity. Ideal for students in Grades 11-12.

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