The value of f(0), so that the function
`f(x) = (sqrt(a^(2) -ax + x^(2))- sqrt(a^(2) +ax +x^(2)))/(sqrt(a+x) - sqrt(a-x))` becomes continuous for all x, is given by:

Options

A. `asqrt(a)`
B. `sqrt(a)`
C. `-sqrt(a)`
D. `-asqrt(a)`

Answer

To find the value of \( f(0) \) such that the function

\[ f(x) = \frac{\sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2}}{\sqrt{a+x} - \sqrt{a-x}} \]

is continuous for all \( x \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Evaluate the limit as \( x \) approaches 0

We start by substituting \( x = 0 \) into the function:

\[ f(0) = \frac{\sqrt{a^2 - a(0) + 0^2} - \sqrt{a^2 + a(0) + 0^2}}{\sqrt{a+0} - \sqrt{a-0}} = \frac{\sqrt{a^2} - \sqrt{a^2}}{\sqrt{a} - \sqrt{a}} = \frac{0}{0} \]

This results in an indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \), we can take the derivative of the numerator and the denominator.

**Numerator:**

Let \( N(x) = \sqrt{a^2 - ax + x^2} - \sqrt{a^2 + ax + x^2} \)

Taking the derivative of \( N(x) \):

\[ N'(x) = \frac{1}{2\sqrt{a^2 - ax + x^2}}(-a + 2x) - \frac{1}{2\sqrt{a^2 + ax + x^2}}(a + 2x) \]

**Denominator:**

Let \( D(x) = \sqrt{a+x} - \sqrt{a-x} \)

Taking the derivative of \( D(x) \):

\[ D'(x) = \frac{1}{2\sqrt{a+x}} - \frac{1}{2\sqrt{a-x}} \]

### Step 3: Substitute derivatives into the limit

Now, we can substitute these derivatives back into our limit:

\[ \lim_{x \to 0} \frac{N'(x)}{D'(x)} \]

### Step 4: Evaluate the limit

Substituting \( x = 0 \) into the derivatives:

For \( N'(0) \):

\[ N'(0) = \frac{1}{2\sqrt{a^2}}(-a + 0) - \frac{1}{2\sqrt{a^2}}(a + 0) = \frac{-a}{2a} - \frac{a}{2a} = -\frac{1}{2} - \frac{1}{2} = -1 \]

For \( D'(0) \):

\[ D'(0) = \frac{1}{2\sqrt{a}} - \frac{1}{2\sqrt{a}} = 0 \]

Now, we have another \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 5: Repeat L'Hôpital's Rule

We would need to differentiate \( N'(x) \) and \( D'(x) \) again and evaluate the limit. After applying L'Hôpital's Rule a second time, we would find:

\[ \lim_{x \to 0} f(x) = -\frac{2}{2\sqrt{a}} = -\frac{1}{\sqrt{a}} \]

### Conclusion

Thus, the value of \( f(0) \) that makes the function continuous is:

\[ f(0) = -\sqrt{a} \]