# How do you find the horizontal asymptote for b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)?

$b \left(x\right) = \frac{{x}^{3} - x + 1}{2 {x}^{4} + {x}^{3} - {x}^{2} - 1} = \frac{{x}^{3} \left(1 - \frac{1}{x} ^ 2 + \frac{1}{x} ^ 3\right)}{{x}^{4} \left(2 + \frac{1}{x} - \frac{1}{x} ^ 2 - \frac{1}{x} ^ 4\right)}$
b(x)=(1-1/x^2+1/x^3)/(x(2+1/x-1/x^2-1/x^4)
${\lim}_{x \to + \infty} \frac{1 - \frac{1}{x} ^ 2 + \frac{1}{x} ^ 3}{x \left(2 + \frac{1}{x} - \frac{1}{x} ^ 2 - \frac{1}{x} ^ 4\right)} = \frac{1}{\infty} = {0}^{+}$
${\lim}_{x \to - \infty} \frac{1 - \frac{1}{x} ^ 2 + \frac{1}{x} ^ 3}{x \left(2 + \frac{1}{x} - \frac{1}{x} ^ 2 - \frac{1}{x} ^ 4\right)} = - \frac{1}{\infty} = {0}^{-}$