Question

If k and b are constant such that lim x approach infinity (kx+b-(x^3+1)/(x^2+1)=0. Find the values of k and b

Answer 1

Combining the terms into one fraction, we have

$kx + b - \dfrac{x^3+1}{x^2+1} = \dfrac{(k-1)x^3 + bx^2 + kx + b - 1}{x^2+1}$

If this converges to 0 as $x\to\infty$, then the degree of the numerator must be smaller than the degree of the denominator.

To ensure this, take $k=1$ and $b=0$. This eliminates the cubic and quadratic terms in the numerator, and we do have

$\displaystyle \lim_{x\to\infty} \frac{x - 1}{x^2 + 1} = \lim_{x\to\infty} \frac{\frac1x - \frac1{x^2}}{1 + \frac1{x^2}} = 0$

Alternatively, we can compute the quotient and remainder of the rational expression.

$\dfrac{x^3+1}{x^2+1} = x - \dfrac{x-1}{x^2+1}$

Then in the limit, we have

$\displaystyle \lim_{x\to\infty} \left(kx + b - x + \frac{x-1}{x^2+1}\right) = (k-1) \lim_{x\to\infty} x + b = 0$

Both terms on the left vanish if $k=1$ and $b=0$.