If `f(x) = (tan(x^(2)-x))/(x), x != 0`, is continuous at x = 0, then f(0)is

Options

A. -1
B. 0
C. 1
D. 2

Answer

To determine the value of \( f(0) \) such that the function \( f(x) = \frac{\tan(x^2 - x)}{x} \) is continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Identify the function and the limit**:
We have \( f(x) = \frac{\tan(x^2 - x)}{x} \) for \( x \neq 0 \). To find \( f(0) \) such that \( f(x) \) is continuous at \( x = 0 \), we need to evaluate:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\tan(x^2 - x)}{x} \]

2. **Evaluate the limit**:
As \( x \to 0 \), both the numerator \( \tan(x^2 - x) \) and the denominator \( x \) approach 0, creating a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and the denominator.

3. **Differentiate the numerator and denominator**:
- The derivative of the numerator \( \tan(x^2 - x) \) is:
\[ \frac{d}{dx}[\tan(x^2 - x)] = \sec^2(x^2 - x) \cdot (2x - 1) \]
- The derivative of the denominator \( x \) is:
\[ \frac{d}{dx}[x] = 1 \]

4. **Apply L'Hôpital's Rule**:
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\tan(x^2 - x)}{x} = \lim_{x \to 0} \frac{\sec^2(x^2 - x) \cdot (2x - 1)}{1} \]

5. **Evaluate the limit as \( x \to 0 \)**:
Substitute \( x = 0 \):
- \( x^2 - x = 0^2 - 0 = 0 \)
- \( \sec^2(0) = 1 \)
- \( 2(0) - 1 = -1 \)

Thus, the limit becomes:
\[ \lim_{x \to 0} \sec^2(x^2 - x) \cdot (2x - 1) = 1 \cdot (-1) = -1 \]

6. **Determine \( f(0) \)**:
For \( f(x) \) to be continuous at \( x = 0 \), we need:
\[ f(0) = \lim_{x \to 0} f(x) = -1 \]

### Conclusion:
Thus, the value of \( f(0) \) that makes the function continuous at \( x = 0 \) is:
\[ \boxed{-1} \]