If `lim_(x->oo) (sqrt(x^2-x+1)-ax-b)=0` then the value of a and b are given by:

Options

A. `a=1,b=(1)/(2)`
B. `a=1 , b=-(1)/(2)`
C. `a=-1,b=(1)/(2)`
D. none of these

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that:

\[ \lim_{x \to \infty} \left( \sqrt{x^2 - x + 1} - ax - b \right) = 0 \]

### Step 1: Rationalize the expression
We start by rationalizing the expression to eliminate the square root. We multiply and divide by the conjugate:

\[ \lim_{x \to \infty} \frac{(\sqrt{x^2 - x + 1} - ax - b)(\sqrt{x^2 - x + 1} + ax + b)}{\sqrt{x^2 - x + 1} + ax + b} \]

This simplifies to:

\[ \lim_{x \to \infty} \frac{x^2 - x + 1 - (ax + b)^2}{\sqrt{x^2 - x + 1} + ax + b} \]

### Step 2: Expand the numerator
Now we expand the numerator:

\[ x^2 - x + 1 - (a^2x^2 + 2abx + b^2) = (1 - a^2)x^2 + (-1 - 2ab)x + (1 - b^2) \]

### Step 3: Analyze the limit
The limit becomes:

\[ \lim_{x \to \infty} \frac{(1 - a^2)x^2 + (-1 - 2ab)x + (1 - b^2)}{\sqrt{x^2 - x + 1} + ax + b} \]

As \( x \to \infty \), the dominant term in the denominator is \( \sqrt{x^2} \), which is \( x \). Thus, we can approximate:

\[ \sqrt{x^2 - x + 1} \sim x \text{ as } x \to \infty \]

So the limit simplifies to:

\[ \lim_{x \to \infty} \frac{(1 - a^2)x^2 + (-1 - 2ab)x + (1 - b^2)}{x + ax + b} \]

### Step 4: Set the coefficients for the limit to be zero
For the limit to equal zero, the coefficient of \( x^2 \) in the numerator must be zero:

\[ 1 - a^2 = 0 \implies a^2 = 1 \implies a = \pm 1 \]

### Step 5: Determine the value of \( b \)
Next, we set the coefficient of \( x \) to zero:

\[ -1 - 2ab = 0 \]

Substituting \( a = 1 \):

\[ -1 - 2(1)b = 0 \implies -1 - 2b = 0 \implies 2b = -1 \implies b = -\frac{1}{2} \]

### Step 6: Verify the case \( a = -1 \)
If \( a = -1 \):

\[ -1 - 2(-1)b = 0 \implies -1 + 2b = 0 \implies 2b = 1 \implies b = \frac{1}{2} \]

However, we need to check if \( a = -1 \) leads to a valid limit. If \( a = -1 \), the denominator becomes problematic as it leads to division by zero. Therefore, we discard \( a = -1 \).

### Final Values
Thus, the values of \( a \) and \( b \) are:

\[ \boxed{a = 1, b = -\frac{1}{2}} \]