If `lim_(xto oo){(x^2+1)/(x+1)-ax-b}=2`, then

Answer

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

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

### Step 1: Simplify the expression inside the limit

We start with the expression:

\[ \frac{x^2 + 1}{x + 1} - ax - b \]

To combine the terms, we can take the common denominator \( x + 1 \):

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

### Step 2: Expand the numerator

Now, we expand the numerator:

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

This simplifies to:

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

### Step 3: Rewrite the limit

Now we can rewrite our limit:

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

### Step 4: Analyze the limit as \( x \to \infty \)

For the limit to exist and equal 2, the coefficient of \( x^2 \) in the numerator must be zero (otherwise the limit would diverge). Therefore, we set:

\[ 1 - a = 0 \implies a = 1 \]

### Step 5: Substitute \( a \) back into the limit

Now substituting \( a = 1 \) into the limit expression:

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

### Step 6: Simplify further

Now, we can factor out \( x \) from the numerator:

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

As \( x \to \infty \), the term \( \frac{1 - b}{x} \) approaches 0, and \( \frac{1}{x} \) also approaches 0. Thus, we have:

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

### Step 7: Set the limit equal to 2

We need this limit to equal 2:

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

### Final Result

Thus, we have found:

\[ a = 1, \quad b = -3 \]