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The given problem involves finding the limit:

\[
\lim_{x \to \infty} \frac{x \cot\left(\frac{5}{x+1}\right)}{1 - x^2}
\]

Let's break this down:

1. As \(x \to \infty\), the expression \(\frac{5}{x+1} \to 0\). Therefore, \(\cot\left(\frac{5}{x+1}\right)\) approaches \(\cot(0)\). However, since \(\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}\) and \(\cos(0) = 1\) and \(\sin(0) = 0\), \(\cot(0)\) is undefined due to division by zero. But \(\cot(\theta) \approx \frac{1}{\theta}\) as \(\theta \to 0\), so \(\cot\left(\frac{5}{x+1}\right) \approx \frac{x+1}{5}\).
2. Therefore, the numerator \(x \cot\left(\frac{5}{x+1}\right)\) can be approximated by \(x \cdot \frac{x+1}{5} = \frac{x(x+1)}{5}\).
3. The denominator \(1 - x^2\) is simply \(1 - x^2\).
4. Putting it together, the expression becomes:
   \[
   \frac{x \cdot \frac{x+1}{5}}{1 - x^2} = \frac{x(x+1)}{5(1 - x^2)}
   \]

As \(x \to \infty\), the leading term of the numerator is \(x^2\) and of the denominator is \(-x^2\). Thus, the limit can be computed as:

\[
\lim_{x \to \infty} \frac{x^2}{-5x^2} = -\frac{1}{5}
\]

So the limit of the given expression as \(x \to \infty\) is \(-\frac{1}{5}\).

Do you want to go into any more details on any step or have any other questions? Here are some possible follow-up questions:

1. How does \(\cot(\theta) \approx \frac{1}{\theta}\) as \(\theta \to 0\) relate to Taylor series?
2. What happens if you alter the constants in the expression?
3. How would the limit change if \(x \to -\infty\) instead?
4. Can this type of problem be generalized for different functions in the numerator?
5. How does the behavior of the trigonometric function affect the overall limit?

**Tip:** When working with limits involving infinity, always look at the leading powers of \(x\) in the numerator and denominator to determine the behavior of the fraction.

lim (x -> ∞) of [x * cot(5/(x+1))] / (1 - x^2)

Solve the limit problem lim (x -> ∞) of [x * cot(5/(x+1))] / (1 - x^2) by approximating cotangent near zero and analyzing the leading powers of x. This problem is ideal for college-level calculus students.

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