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The problem asks to calculate the limit:

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

Let's break it into two parts and compute each one separately.

### 1. First term: \(\frac{n^3 - 1}{n^3}\)
\[
\frac{n^3 - 1}{n^3} = 1 - \frac{1}{n^3}
\]
As \(n \to \infty\), \(\frac{1}{n^3} \to 0\), so the limit of this term is:
\[
1 - 0 = 1
\]

### 2. Second term: \(\frac{1}{n+1} - \frac{1}{1-n}\)
We need to simplify this difference of fractions:
\[
\frac{1}{n+1} - \frac{1}{1-n} = \frac{(1-n) - (n+1)}{(n+1)(1-n)} = \frac{1 - n - n - 1}{(n+1)(1-n)} = \frac{-2n}{(n+1)(1-n)}
\]
Now, simplify the denominator:
\[
(n+1)(1-n) = -n^2 + 1
\]
Thus, the second term becomes:
\[
\frac{-2n}{-n^2 + 1}
\]
For large \(n\), this expression behaves as:
\[
\frac{-2n}{-n^2 + 1} \approx \frac{-2n}{-n^2} = \frac{2}{n}
\]
As \(n \to \infty\), \(\frac{2}{n} \to 0\).

### Final result:
Combining both parts:
\[
1 + 0 = 1
\]

Thus, the limit is:

\[
\boxed{1}
\]

Would you like more details or explanations on any part of this? Here are 5 related questions:

1. How do limits involving infinity work with higher-degree polynomials?
2. What are common techniques for simplifying limits involving rational expressions?
3. Why does \(\frac{1}{n^3} \to 0\) as \(n \to \infty\)?
4. How do you handle limits involving differences of fractions?
5. Can you explain the role of the degree of polynomials in determining limits at infinity?

**Tip:** When working with limits, always simplify complex expressions step by step, and remember that terms that approach zero as \(n \to \infty\) often simplify calculations.

Berechnen Sie den Grenzwert: lim_{n → ∞} (n^3 - 1) / n^3 + (1/(n+1) - 1/(1-n))

This detailed explanation shows how to calculate the limit lim_{n → ∞} (n^3 - 1) / n^3 + (1/(n+1) - 1/(1-n)). Step-by-step breakdown includes algebraic manipulation of polynomials and rational expressions.

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