If `lim_(xrarroo) {(x^3+1)/(x^2+1)-(ax+b)}=2`, then

Options

A. `a=1,b=1`
B. `a=1,b=2`
C. `a=1,b=-2`
D. none of these

Answer

To solve the limit problem given by

\[ \lim_{x \to \infty} \left( \frac{x^3 + 1}{x^2 + 1} - (ax + b) \right) = 2, \]

we will follow these steps:

### Step 1: Rewrite the limit expression
We start with the limit expression:

\[ \lim_{x \to \infty} \left( \frac{x^3 + 1}{x^2 + 1} - (ax + b) \right). \]

### Step 2: Combine the fractions
To combine the terms, we need a common denominator. The common denominator is \(x^2 + 1\):

\[ \frac{x^3 + 1 - (ax + b)(x^2 + 1)}{x^2 + 1}. \]

### Step 3: Expand the numerator
Now, we expand the numerator:

\[ x^3 + 1 - (ax^3 + bx^2 + ax + b) = x^3 + 1 - ax^3 - bx^2 - ax - b. \]

This simplifies to:

\[ (1 - a)x^3 - bx^2 - ax + (1 - b). \]

### Step 4: Set up the limit
Now we have:

\[ \lim_{x \to \infty} \frac{(1 - a)x^3 - bx^2 - ax + (1 - b)}{x^2 + 1} = 2. \]

### Step 5: Analyze the leading terms
For the limit to exist and be finite as \(x\) approaches infinity, the highest degree terms in the numerator and denominator must balance. The highest degree term in the denominator is \(x^2\), so we need the coefficient of \(x^3\) in the numerator to be zero:

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

### Step 6: Substitute \(a\) back into the limit
Now substituting \(a = 1\) into the expression, we have:

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

### Step 7: Simplify the limit
Now, we can factor out \(x^2\) from the numerator:

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

### Step 8: Evaluate the limit
As \(x\) approaches infinity, the terms \(\frac{1}{x}\) and \(\frac{1}{x^2}\) approach zero:

\[ \lim_{x \to \infty} \frac{-b}{1} = -b. \]

### Step 9: Set the limit equal to 2
We know from the problem statement that this limit equals 2:

\[ -b = 2 \implies b = -2. \]

### Conclusion
Thus, we have found the values:

\[ a = 1 \quad \text{and} \quad b = -2. \]