If `lim_(xto oo)((x^(2)+1)/(x+1)-ax-b)=oo` find a and b.

Answer

To solve the limit problem, we need to analyze the expression given:

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

### Step 1: Simplify the expression

First, we simplify the fraction:

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

As \(x\) approaches infinity, the term \(\frac{1}{x + 1}\) approaches 0. Therefore, we focus on the leading term:

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

### Step 2: Rewrite the limit expression

Now we can rewrite the limit:

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

This simplifies to:

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

### Step 3: Determine conditions for the limit to be infinity

For the limit to equal infinity, the coefficient of \(x\) must be positive:

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

### Step 4: Analyze the constant term

The term \(-b\) does not affect the limit as \(x\) approaches infinity, so \(b\) can be any real number.

### Conclusion

Thus, we conclude:

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

### Final Answer

\[ \text{a} < 1, \quad \text{b} \in \mathbb{R} \]