Is $ \sqrt{2} $ the limit of a sequence of numbers of the form $ \sqrt[3]{n} - \sqrt[3]{m} $, where $ n, m = 0, 1, 2, \cdots $.
1. **Define the sequence:**
   For each \( k \in \mathbb{N} \), let \( n = \lfloor (k + \sqrt{2})^3 \rfloor \) and \( m = k^3 \). We need to show that the sequence \( \sqrt[3]{n} - \sqrt[3]{m} \) converges to \( \sqrt{2} \).

2. **Upper bound:**
   \[
   \sqrt[3]{n} - \sqrt[3]{m} = \sqrt[3]{\lfloor (k + \sqrt{2})^3 \rfloor} - k < \sqrt[3]{(k + \sqrt{2})^3} - k = \sqrt{2}
   \]
   This inequality holds because \( \lfloor (k + \sqrt{2})^3 \rfloor \leq (k + \sqrt{2})^3 \).

3. **Lower bound:**
   \[
   \sqrt[3]{n} - \sqrt[3]{m} = \sqrt[3]{\lfloor (k + \sqrt{2})^3 \rfloor} - k > \sqrt[3]{(k + \sqrt{2})^3 - 1} - k
   \]
   To analyze this, we use the binomial expansion:
   \[
   (k + \sqrt{2})^3 = k^3 + 3k^2\sqrt{2} + 3k(\sqrt{2})^2 + (\sqrt{2})^3 = k^3 + 3k^2\sqrt{2} + 6k + 2\sqrt{2}
   \]
   Therefore,
   \[
   (k + \sqrt{2})^3 - 1 = k^3 + 3k^2\sqrt{2} + 6k + 2\sqrt{2} - 1
   \]

4. **Simplify the lower bound expression:**
   \[
   \sqrt[3]{(k + \sqrt{2})^3 - 1} - k = \frac{(k + \sqrt{2})^3 - 1 - k^3}{\sqrt[3]{((k + \sqrt{2})^3 - 1)^2} + k\sqrt[3]{(k + \sqrt{2})^3 - 1} + k^2}
   \]
   Substituting the expanded form:
   \[
   \frac{3k^2\sqrt{2} + 6k + 2\sqrt{2} - 1}{\sqrt[3]{((k + \sqrt{2})^3 - 1)^2} + k\sqrt[3]{(k + \sqrt{2})^3 - 1} + k^2}
   \]

5. **Analyze the limit:**
   As \( k \to \infty \), the dominant term in the numerator is \( 3k^2\sqrt{2} \) and in the denominator, the dominant term is \( 3k^2 \). Thus,
   \[
   \frac{3k^2\sqrt{2} + 6k + 2\sqrt{2} - 1}{\sqrt[3]{((k + \sqrt{2})^3 - 1)^2} + k\sqrt[3]{(k + \sqrt{2})^3 - 1} + k^2} \approx \frac{3k^2\sqrt{2}}{3k^2} = \sqrt{2}
   \]

6. **Apply the squeeze theorem:**
   Since \( \sqrt[3]{n} - \sqrt[3]{m} \) is squeezed between \( \sqrt{2} \) and a term that approaches \( \sqrt{2} \) as \( k \to \infty \), by the squeeze theorem, we conclude that:
   \[
   \lim_{k \to \infty} \left( \sqrt[3]{\lfloor (k + \sqrt{2})^3 \rfloor} - \sqrt[3]{k^3} \right) = \sqrt{2}
   \]

The final answer is \( \boxed{ \sqrt{2} } \)