Find the values of a and b if lim (x->inf) [ (x3+1)/(x2+1) - (ax+b) ] = 2.

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

Expand the product $(ax+b)(x^2+1)$:

$$(ax+b)(x^2+1) = ax(x^2+1) + b(x^2+1) = ax^3 + ax + bx^2 + b$$

Substitute this back into the numerator:

$$x^3+1 - (ax^3 + bx^2 + ax + b) = x^3+1 - ax^3 - bx^2 - ax - b$$

Rearrange the terms by powers of $x$:

$$(1-a)x^3 - bx^2 - ax + (1-b)$$

So the limit becomes:

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

For the limit of a rational function as $x \to \infty$ to be a finite non-zero value, the degree of the numerator must be equal to the degree of the denominator.
The degree of the denominator is 2. The current degree of the numerator is 3 (due to the $x^3$ term).
For the limit to be finite, the coefficient of $x^3$ in the numerator must be zero.

$$1-a = 0 \implies a = 1$$

With $a=1$, the expression becomes:

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

To evaluate this limit, divide every term in the numerator and denominator by the highest power of $x$ in the denominator, which is $x^2$:

$$\lim_{x \to \infty} \left( \frac{\frac{-bx^2}{x^2} - \frac{x}{x^2} + \frac{1-b}{x^2}}{\frac{x^2}{x^2} + \frac{1}{x^2}} \right) = \lim_{x \to \infty} \left( \frac{-b - \frac{1}{x} + \frac{1-b}{x^2}}{1 + \frac{1}{x^2}} \right)$$

As $x \to \infty$, the terms $\frac{1}{x}$ and $\frac{1}{x^2}$ approach 0.

$$\frac{-b - 0 + 0}{1 + 0} = -b$$

We found $a=1$ and $b=-2$. This matches option (C).

The final answer is $\text{C) a = 1, b = -2}$.