The value of `lim_(ntooo) [root(3)((n+1)^(2))-root(3)((n-1)^(2))]` is __________.

Options

A. Equals `(1)/(sqrt(2))`
B. Does not exist
C. Equals `sqrt(2)`
D. Equals -sqrt(2)`

Explanation

To solve the limit \( \lim_{n \to \infty} \left( \sqrt[3]{(n+1)^2} - \sqrt[3]{(n-1)^2} \right) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{n \to \infty} \left( \sqrt[3]{(n+1)^2} - \sqrt[3]{(n-1)^2} \right) \]

### Step 2: Factor out the dominant term
As \( n \to \infty \), both \( (n+1)^2 \) and \( (n-1)^2 \) can be approximated by \( n^2 \). Thus, we can factor \( n^2 \) out of the cube roots:
\[ \sqrt[3]{(n+1)^2} = \sqrt[3]{n^2 \left(1 + \frac{1}{n}\right)^2} = n^{2/3} \sqrt[3]{\left(1 + \frac{1}{n}\right)^2} \]
\[ \sqrt[3]{(n-1)^2} = \sqrt[3]{n^2 \left(1 - \frac{1}{n}\right)^2} = n^{2/3} \sqrt[3]{\left(1 - \frac{1}{n}\right)^2} \]

### Step 3: Substitute back into the limit
Now substituting these back into the limit:
\[ \lim_{n \to \infty} \left( n^{2/3} \sqrt[3]{\left(1 + \frac{1}{n}\right)^2} - n^{2/3} \sqrt[3]{\left(1 - \frac{1}{n}\right)^2} \right) \]
Factoring out \( n^{2/3} \):
\[ = n^{2/3} \lim_{n \to \infty} \left( \sqrt[3]{\left(1 + \frac{1}{n}\right)^2} - \sqrt[3]{\left(1 - \frac{1}{n}\right)^2} \right) \]

### Step 4: Evaluate the limit inside the parentheses
Next, we evaluate the limit inside the parentheses:
\[ \sqrt[3]{\left(1 + \frac{1}{n}\right)^2} \approx 1 + \frac{2}{3n} \quad \text{and} \quad \sqrt[3]{\left(1 - \frac{1}{n}\right)^2} \approx 1 - \frac{2}{3n} \]
Thus,
\[ \sqrt[3]{\left(1 + \frac{1}{n}\right)^2} - \sqrt[3]{\left(1 - \frac{1}{n}\right)^2} \approx \left(1 + \frac{2}{3n}\right) - \left(1 - \frac{2}{3n}\right) = \frac{4}{3n} \]

### Step 5: Substitute back into the limit
Now substituting this back:
\[ n^{2/3} \cdot \frac{4}{3n} = \frac{4}{3} n^{-1/3} \]

### Step 6: Evaluate the final limit
Finally, we take the limit as \( n \to \infty \):
\[ \lim_{n \to \infty} \frac{4}{3} n^{-1/3} = 0 \]

### Conclusion
Thus, the value of the limit is:
\[ \boxed{0} \]