`lim_(x to -infty)sqrt(x^(2)-x+1)-ax + b=0`, then (a,b) is:

Options

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

Answer

To solve the limit problem \( \lim_{x \to -\infty} \left( \sqrt{x^2 - x + 1} - ax + b \right) = 0 \), we will follow these steps:

### Step 1: Analyze the limit
We start by rewriting the limit:
\[ L = \lim_{x \to -\infty} \left( \sqrt{x^2 - x + 1} - ax + b \right) \]
We want \( L = 0 \).

### Step 2: Simplify the square root
For large negative \( x \), we can approximate the expression inside the square root:
\[ \sqrt{x^2 - x + 1} \approx \sqrt{x^2} = |x| = -x \quad (\text{since } x \text{ is negative}) \]
Thus, we can write:
\[ \sqrt{x^2 - x + 1} \approx -x \text{ as } x \to -\infty \]

### Step 3: Substitute the approximation into the limit
Now substituting this approximation into our limit:
\[ L \approx \lim_{x \to -\infty} \left( -x - ax + b \right) = \lim_{x \to -\infty} \left( -(1 + a)x + b \right) \]

### Step 4: Analyze the resulting expression
For the limit to equal zero as \( x \to -\infty \), the coefficient of \( x \) must be zero:
\[ -(1 + a) = 0 \implies 1 + a = 0 \implies a = -1 \]

### Step 5: Substitute \( a \) back into the limit
Now substituting \( a = -1 \):
\[ L = \lim_{x \to -\infty} \left( -(-1)x + b \right) = \lim_{x \to -\infty} \left( x + b \right) \]
For this limit to be zero, we need:
\[ b = -\infty \text{ (which is not a valid constant)} \]
So, we need to check for the next condition.

### Step 6: Find \( b \) when \( a = -1 \)
We have:
\[ L = \lim_{x \to -\infty} \left( x + b \right) = 0 \implies b = -1 \]

### Final Result
Thus, we have found:
\[ (a, b) = (-1, -1) \]

### Conclusion
The values of \( a \) and \( b \) that satisfy the limit condition are:
\[ \boxed{(-1, -1)} \]