If `lim_(xrarroo) {ax-(x^2+1)/(x+1)}=b` , a finite number, then

Options

A. `a=1, b=1`
B. `a=0,b=1`
C. `a=-1,b=1`
D. `b=-1,b=-1`

Answer

To solve the limit problem given by the expression \( \lim_{x \to \infty} \left( ax - \frac{x^2 + 1}{x + 1} \right) = b \), where \( a \) and \( b \) are finite numbers, we can follow these steps:

### Step-by-Step Solution

1. **Rewrite the expression**:
We start with the limit:
\[ \lim_{x \to \infty} \left( ax - \frac{x^2 + 1}{x + 1} \right) \]

2. **Combine the terms**:
To combine the terms, we can rewrite the expression as:
\[ ax - \frac{x^2 + 1}{x + 1} = ax - \left( \frac{x^2}{x + 1} + \frac{1}{x + 1} \right) \]
This simplifies to:
\[ ax - \frac{x^2}{x + 1} - \frac{1}{x + 1} \]

3. **Find a common denominator**:
The common denominator for the terms is \( x + 1 \):
\[ = \frac{(ax)(x + 1) - (x^2 + 1)}{x + 1} \]

4. **Expand the numerator**:
Expanding the numerator gives:
\[ = \frac{ax^2 + ax - x^2 - 1}{x + 1} \]
This can be rearranged as:
\[ = \frac{(a - 1)x^2 + ax - 1}{x + 1} \]

5. **Analyze the limit**:
Now we take the limit as \( x \to \infty \):
\[ \lim_{x \to \infty} \frac{(a - 1)x^2 + ax - 1}{x + 1} \]
For this limit to be finite, the degree of the numerator must equal the degree of the denominator. The denominator is of degree 1, so the coefficient of \( x^2 \) in the numerator must be zero:
\[ a - 1 = 0 \implies a = 1 \]

6. **Substitute \( a \) back into the limit**:
Substitute \( a = 1 \) back into the limit:
\[ \lim_{x \to \infty} \frac{(1 - 1)x^2 + (1)x - 1}{x + 1} = \lim_{x \to \infty} \frac{0 + x - 1}{x + 1} \]
This simplifies to:
\[ = \lim_{x \to \infty} \frac{x - 1}{x + 1} \]

7. **Simplify the limit**:
Factor out \( x \):
\[ = \lim_{x \to \infty} \frac{x(1 - \frac{1}{x})}{x(1 + \frac{1}{x})} = \lim_{x \to \infty} \frac{1 - \frac{1}{x}}{1 + \frac{1}{x}} = \frac{1 - 0}{1 + 0} = 1 \]

8. **Conclusion**:
Thus, we find that \( b = 1 \).

### Final Result:
The values are \( a = 1 \) and \( b = 1 \).