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 $\tan^{-1}(\pm \sqrt{3})$.