The value of `lim_(x->1) y^3/(x^3-y^2-1)` as `(x, y)-> (1, 0)` along the line `y = x -1` is

Options

A. `1`
B. `oo`
C. `0`
D. None of these

Answer

To solve the limit problem, we need to evaluate the expression:

\[ \lim_{x \to 1} \frac{y^3}{x^3 - y^2 - 1} \]

as \((x, y) \to (1, 0)\) along the line \(y = x - 1\).

### Step 1: Substitute \(y\) in terms of \(x\)
Since we are given that \(y = x - 1\), we can substitute this into the limit expression.

\[ y^3 = (x - 1)^3 \]

### Step 2: Rewrite the limit
Now we can rewrite the limit:

\[ \lim_{x \to 1} \frac{(x - 1)^3}{x^3 - (x - 1)^2 - 1} \]

### Step 3: Simplify the denominator
Next, we simplify the denominator:

\[ (x - 1)^2 = x^2 - 2x + 1 \]

Thus,

\[ x^3 - (x - 1)^2 - 1 = x^3 - (x^2 - 2x + 1) - 1 = x^3 - x^2 + 2x - 1 - 1 = x^3 - x^2 + 2x - 2 \]

### Step 4: Rewrite the limit again
Now the limit becomes:

\[ \lim_{x \to 1} \frac{(x - 1)^3}{x^3 - x^2 + 2x - 2} \]

### Step 5: Evaluate the limit
Now we substitute \(x = 1\):

\[ \frac{(1 - 1)^3}{1^3 - 1^2 + 2(1) - 2} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 6: Apply L'Hôpital's Rule
Differentiate the numerator and the denominator:

- The derivative of the numerator \((x - 1)^3\) is \(3(x - 1)^2\).
- The derivative of the denominator \(x^3 - x^2 + 2x - 2\) is \(3x^2 - 2x + 2\).

Thus, we have:

\[ \lim_{x \to 1} \frac{3(x - 1)^2}{3x^2 - 2x + 2} \]

### Step 7: Substitute \(x = 1\) again
Now we substitute \(x = 1\) again:

\[ \frac{3(1 - 1)^2}{3(1)^2 - 2(1) + 2} = \frac{0}{3 - 2 + 2} = \frac{0}{3} = 0 \]

### Final Answer
Thus, the value of the limit is:

\[ \boxed{0} \]