f(x) is differentiable, increasing functions, then `lim_(x to 0)(f(x^(2))-f(x))/(f(x)-f(0))` is equal to:

Options

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

Answer

To solve the limit \( \lim_{x \to 0} \frac{f(x^2) - f(x)}{f(x) - f(0)} \), we will follow these steps:

### Step 1: Identify the form of the limit
As \( x \to 0 \), both the numerator \( f(x^2) - f(x) \) and the denominator \( f(x) - f(0) \) approach \( 0 \). This gives us a \( \frac{0}{0} \) indeterminate form.

**Hint:** When you encounter a \( \frac{0}{0} \) form, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we can take the derivative of the numerator and the derivative of the denominator:

\[ \lim_{x \to 0} \frac{f(x^2) - f(x)}{f(x) - f(0)} = \lim_{x \to 0} \frac{(f(x^2))' - (f(x))'}{(f(x))'} \]

### Step 3: Differentiate the numerator and denominator
Using the chain rule for the derivative of \( f(x^2) \):

- The derivative of \( f(x^2) \) is \( f'(x^2) \cdot 2x \).
- The derivative of \( f(x) \) is \( f'(x) \).

Thus, the limit becomes:

\[ \lim_{x \to 0} \frac{f'(x^2) \cdot 2x - f'(x)}{f'(x)} \]

### Step 4: Substitute \( x = 0 \)
Now, we substitute \( x = 0 \):

- \( f'(x^2) \) approaches \( f'(0) \) as \( x \to 0 \).
- The term \( 2x \) approaches \( 0 \).
- Therefore, the numerator becomes \( f'(0) \cdot 0 - f'(0) = -f'(0) \).
- The denominator becomes \( f'(0) \).

Thus, the limit simplifies to:

\[ \lim_{x \to 0} \frac{-f'(0)}{f'(0)} = -1 \]

### Final Answer
The value of the limit is:

\[ \boxed{-1} \]

---