Asymptote Math Example 2

Solution

1. 1
   Since the numerator degree (2) is exactly one more than the denominator degree (1), an oblique asymptote exists. Perform polynomial long division: $x^2 + x - 1 \div (x-1)$.
2. 2
   First term: $x^2 \div x = x$. Multiply: $x(x-1) = x^2-x$. Subtract: $(x^2+x-1)-(x^2-x) = 2x-1$.
3. 3
   Second term: $2x \div x = 2$. Multiply: $2(x-1)=2x-2$. Subtract: $(2x-1)-(2x-2)=1$. So $g(x) = x + 2 + \frac{1}{x-1}$. As $x \to \pm\infty$, the remainder $\frac{1}{x-1} \to 0$, giving the oblique asymptote $y = x + 2$.