Combining the terms into one fraction, we have
[tex]kx + b - \dfrac{x^3+1}{x^2+1} = \dfrac{(k-1)x^3 + bx^2 + kx + b - 1}{x^2+1}[/tex]
If this converges to 0 as [tex]x\to\infty[/tex], then the degree of the numerator must be smaller than the degree of the denominator.
To ensure this, take [tex]k=1[/tex] and [tex]b=0[/tex]. This eliminates the cubic and quadratic terms in the numerator, and we do have
[tex]\displaystyle \lim_{x\to\infty} \frac{x - 1}{x^2 + 1} = \lim_{x\to\infty} \frac{\frac1x - \frac1{x^2}}{1 + \frac1{x^2}} = 0[/tex]
Alternatively, we can compute the quotient and remainder of the rational expression.
[tex]\dfrac{x^3+1}{x^2+1} = x - \dfrac{x-1}{x^2+1}[/tex]
Then in the limit, we have
[tex]\displaystyle \lim_{x\to\infty} \left(kx + b - x + \frac{x-1}{x^2+1}\right) = (k-1) \lim_{x\to\infty} x + b = 0[/tex]
Both terms on the left vanish if [tex]k=1[/tex] and [tex]b=0[/tex].
