https://file.aiquickdraw.com/mathbot/file/17271926740260mhnwx2o.jpg

The image asks to solve the following limit problem:

\[
\lim_{x \to \infty} \left( \frac{1}{2} - x \cdot \tan\left(\frac{8}{x}\right) \right)
\]

Let's break it down and solve step by step.

### Step 1: Approximate \(\tan\left(\frac{8}{x}\right)\) as \(x \to \infty\)
For very large values of \(x\), \(\frac{8}{x}\) approaches zero. We can use the approximation for small angles:
\[
\tan\left(\frac{8}{x}\right) \approx \frac{8}{x}
\]
This approximation holds because \(\tan(\theta) \approx \theta\) for \(\theta\) close to zero.

### Step 2: Substitute the approximation into the limit expression
Now substitute \(\tan\left(\frac{8}{x}\right) \approx \frac{8}{x}\) into the limit expression:
\[
\lim_{x \to \infty} \left( \frac{1}{2} - x \cdot \frac{8}{x} \right)
\]
Simplifying the expression:
\[
= \lim_{x \to \infty} \left( \frac{1}{2} - 8 \right)
= \frac{1}{2} - 8 = -\frac{15}{2}
\]

### Final Answer:
\[
\lim_{x \to \infty} \left( \frac{1}{2} - x \cdot \tan\left(\frac{8}{x}\right) \right) = -\frac{15}{2}
\]

Would you like further explanation or details on any part of the solution?

Here are 5 related questions to explore this topic further:
1. How does the small-angle approximation for trigonometric functions work?
2. What other methods can be used to evaluate limits involving trigonometric functions?
3. Can you apply L'Hopital's Rule to solve this limit problem?
4. How does the behavior of \(\tan(x)\) change for large and small values of \(x\)?
5. What are common approximations used for limits approaching infinity?

**Tip:** When dealing with trigonometric limits as \(x\) approaches infinity, small-angle approximations often simplify the process.

Tentukan nilai dari \( \lim_{x \to \infty} \left( \frac{1}{2} - x \cdot \tan\left( \frac{8}{x} \right) \right) \).

Discover how to solve the limit problem \( \lim_{x \to \infty} \left( \frac{1}{2} - x \cdot \tan\left( \frac{8}{x} \right) \right) \) using the small-angle approximation for trigonometric functions. This detailed solution involves calculus and is suitable for Grades 11-12 or college-level students.

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