If `"lt"_(x to infty) ((x^(3)+1)/(x^(2)+1) -ax-b)=2`, then (a,b) =

Options

A. (1,-2)
B. (1,2)
C. (1,1)
D. none

Answer

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

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

### Step 1: Simplify the expression

We start by simplifying the expression inside the limit:

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

### Step 2: Find the leading term as \( x \to \infty \)

As \( x \) approaches infinity, we can focus on the leading terms in both the numerator and the denominator:

\[ \frac{x^3 + 1}{x^2 + 1} \approx \frac{x^3}{x^2} = x \]

Thus, we rewrite the limit:

\[ \lim_{x \to \infty} \left( x - ax - b \right) \]

### Step 3: Combine like terms

Now, we combine the terms:

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

### Step 4: Analyze the limit

For the limit to be finite (specifically equal to 2), the coefficient of \( x \) must be zero:

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

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

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

\[ \lim_{x \to \infty} \left( (1 - 1)x - b \right) = -b \]

### Step 6: Set the limit equal to 2

We set this equal to 2:

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

### Conclusion

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

\[ (a, b) = (1, -2) \]