If `a gt 0` and `lim_(xrarr oo) {sqrt(x^2+x+1)-(ax+b)}=0`, then `(a,b)` lies on the line.

Options

A. `x-y+3=0`
B. `3x+4y-5=0`
C. `x+6y+2=0`
D. `x+2y+3=0`

Explanation

To solve the problem, we need to analyze the limit given and find the values of \(a\) and \(b\) such that the limit approaches zero.

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

### Step 1: Simplify the expression inside the limit
We can start by rewriting the expression:
\[ \sqrt{x^2 + x + 1} - (ax + b) \]
As \(x\) approaches infinity, the dominant term inside the square root is \(x^2\). Thus, we can factor \(x^2\) out:
\[ \sqrt{x^2(1 + \frac{1}{x} + \frac{1}{x^2})} = x\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} \]
So we rewrite the limit:
\[ \lim_{x \to \infty} \left( x\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - (ax + b) \right) \]

### Step 2: Factor out \(x\)
Now we can factor \(x\) from the limit:
\[ \lim_{x \to \infty} \left( x \left( \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - a \right) - b \right) \]

### Step 3: Analyze the limit
For the limit to equal zero as \(x\) approaches infinity, the term inside the parentheses must approach zero. Therefore:
\[ \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} \to 1 \quad \text{as } x \to \infty \]
Thus, we need:
\[ 1 - a = 0 \implies a = 1 \]

### Step 4: Substitute \(a\) back into the limit
Now substituting \(a = 1\) into the limit:
\[ \lim_{x \to \infty} \left( x(1 - 1) - b \right) = -b \]
For this limit to equal zero, we must have:
\[ -b = 0 \implies b = 0 \]

### Step 5: Conclusion
Thus, we find that:
\[ (a, b) = (1, 0) \]

### Step 6: Determine the line
Now we need to check if the point \((1, 0)\) lies on the given lines.

1. For the line \(x - y + 3 = 0\):
\[ 1 - 0 + 3 = 4 \quad \text{(not equal to 0)} \]
2. For the line \(3x + 4y - 5 = 0\):
\[ 3(1) + 4(0) - 5 = 3 - 5 = -2 \quad \text{(not equal to 0)} \]

### Final Result
The point \((1, 0)\) does not lie on either of the lines provided in the options.