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The problem is to find the limit of the given expression as \(x \to \infty\):

\[
\lim_{{x \to \infty}} \tan^{-1} \left( \frac{1 - \sqrt{3}x}{3 + x} \right)
\]

### Step-by-step Solution:

1. **Simplify the Expression Inside the Arctan Function:**

   Consider the fraction:

   \[
   \frac{1 - \sqrt{3}x}{3 + x}
   \]

   As \(x \to \infty\), the terms involving \(x\) will dominate over the constant terms. Thus, we can divide both the numerator and the denominator by \(x\):

   \[
   \frac{\frac{1 - \sqrt{3}x}{x}}{\frac{3 + x}{x}} = \frac{\frac{1}{x} - \sqrt{3}}{\frac{3}{x} + 1}
   \]

2. **Evaluate the Limit as \(x \to \infty\):**

   As \(x \to \infty\), \(\frac{1}{x} \to 0\) and \(\frac{3}{x} \to 0\). Thus, the expression simplifies to:

   \[
   \frac{0 - \sqrt{3}}{0 + 1} = -\sqrt{3}
   \]

   So, the expression inside the arctan function converges to \(-\sqrt{3}\).

3. **Apply the Arctan Function and Find the Limit:**

   Now, we need to compute:

   \[
   \lim_{{x \to \infty}} \tan^{-1}(-\sqrt{3})
   \]

   Since \(\tan^{-1}(-\sqrt{3})\) is a constant, the limit is simply this constant. We know that:

   \[
   \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}
   \]

4. **Conclusion:**

   Therefore, the limit is:

   \[
   \lim_{{x \to \infty}} \tan^{-1} \left( \frac{1 - \sqrt{3}x}{3 + x} \right) = -\frac{\pi}{3}
   \]

### Final Answer:

\[
\boxed{-\frac{\pi}{3}}
\]

Would you like more details on any step or have any questions?

### Related Questions:

1. What is the limit of \(\lim_{x \to -\infty} \tan^{-1} \left( \frac{1 - \sqrt{3}x}{3 + x} \right)\)?
2. How can we find the limit of \(\lim_{x \to \infty} \arcsin \left( \frac{1 - \sqrt{3}x}{3 + x} \right)\)?
3. What is the limit of \(\lim_{x \to 0} \tan^{-1} \left( \frac{1 - \sqrt{3}x}{3 + x} \right)\)?
4. Can you find the limit \(\lim_{x \to \infty} \frac{1 - \sqrt{3}x}{3 + x}\) without using the arctan function?
5. What is the behavior of the function \(f(x) = \tan^{-1} \left( \frac{1 - \sqrt{3}x}{3 + x} \right)\) as \(x \to \infty\)?

### Tip:

Remember that the limit of an inverse trigonometric function often involves recognizing standard values like \(\tan^{-1}(\pm 1)\) or \(\tan^{-1}(\pm \sqrt{3})\).

Find the limit: lim_{x→∞} arctan((1 - √3x)/(3 + x))

Find the limit of arctan((1 - √3x)/(3 + x)) as x approaches infinity. This solution covers key concepts in limits and inverse trigonometric functions. Suitable for students in Grades 11-12 or university level.

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