If `Lt_(x to infty) ((x^(2)-1)/(x+1) - ax-b)=2`, then (a,b) =

Options

A. (1,3)
B. `(-1,3)`
C. `(1,-3)`
D. none

Answer

To solve the given limit problem, we start with the expression:

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

### Step 1: Rewrite the limit expression
We can rewrite the limit expression as follows:

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

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

\[ x^2 - 1 - (ax + b)(x + 1) = x^2 - 1 - (ax^2 + ax + bx + b) \]

This simplifies to:

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

### Step 3: Substitute into the limit
Now we substitute this back into the limit:

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

### Step 4: Factor out the highest power of x
As \(x\) approaches infinity, we can factor out \(x^2\) from the numerator and \(x\) from the denominator:

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

This simplifies to:

\[ \lim_{x \to \infty} \frac{(1 - a)x}{1 + \frac{1}{x}} = (1 - a)x \]

### Step 5: Set the limit equal to 2
For the limit to equal 2, we need the coefficient of \(x\) to be zero (to avoid infinity) and the constant term to equal 2. Therefore, we set:

1. \(1 - a = 0\) (to eliminate the \(x\) term)
2. The constant term must equal 2.

From \(1 - a = 0\), we find:

\[ a = 1 \]

### Step 6: Substitute \(a\) back to find \(b\)
Now substituting \(a = 1\) into the expression for the constant term:

\[ -(1 + b) = 2 \implies -1 - b = 2 \implies -b = 3 \implies b = -3 \]

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

\[ (a, b) = (1, -3) \]