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

Options

A. [1,2]
B. (2,3}
C. (3,4)
D. none

Answer

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

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

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

First, we simplify the term \(\frac{x^2 + 1}{x + 1}\):

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

As \(x\) approaches infinity, the dominant term in the numerator is \(x^2\) and in the denominator is \(x\). Thus, we can approximate:

\[ \frac{x^2 + 1}{x + 1} \approx \frac{x^2}{x} = x \quad \text{(as } x \to \infty\text{)} \]

### Step 2: Substitute the approximation back into the limit

Now, substituting this back into our limit expression gives:

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

### Step 3: Analyze the limit condition

For the limit to approach infinity, the coefficient of \(x\) must be positive. Therefore, we need:

\[ 1 - a > 0 \implies a < 1 \]

### Step 4: Consider the constant term

The term \(-b\) does not affect the limit approaching infinity since it is a constant. Thus, \(b\) can be any real number.

### Conclusion

From the analysis above, we conclude that:

- \(a < 1\)
- \(b\) can be any real number.

Thus, the solution set for \((a, b)\) is:

\[ (a, b) = (a < 1, b \in \mathbb{R}) \]