To find the limit lim x → ± ∞ x 2 − 8 x − 4 {\displaystyle {\underset {x\to \pm \infty }{\lim }}{\frac {\sqrt {x^{2}-8}}{x-4}}} , the first step is to factor out the highest degree term from numerator and denominator, so we have
lim x → ± ∞ x 2 − 8 x − 4 = lim x → ± ∞ x 2 ( 1 − 8 x 2 ) x ( 1 − 4 x ) = lim x → ± ∞ x 2 1 − 8 x 2 x ( 1 − 4 x ) = lim x → ± ∞ | x | 1 − 8 x 2 x ( 1 − 4 x ) ∗ Note : x 2 = | x | = { lim x → + ∞ x 1 − 8 x 2 x ( 1 − 4 x ) = lim x → + ∞ 1 − 8 x 2 ( 1 − 4 x ) = 1 lim x → − ∞ − x 1 − 8 x 2 x ( 1 − 4 x ) = lim x → + ∞ − 1 − 8 x 2 ( 1 − 4 x ) = − 1 {\displaystyle {\begin{aligned}{\underset {x\to \pm \infty }{\lim }}{\frac {\sqrt {x^{2}-8}}{x-4}}&={\underset {x\to \pm \infty }{\lim }}{\frac {\sqrt {x^{2}(1-{\frac {8}{x^{2}}})}}{x(1-{\frac {4}{x}})}}\\&={\underset {x\to \pm \infty }{\lim }}{\frac {{\sqrt {x^{2}}}{\sqrt {1-{\frac {8}{x^{2}}}}}}{x(1-{\frac {4}{x}})}}\\&={\underset {x\to \pm \infty }{\lim }}{\frac {|x|{\sqrt {1-{\frac {8}{x^{2}}}}}}{x(1-{\frac {4}{x}})}}\qquad \qquad *{\text{Note}}:{\sqrt {x^{2}}}=|x|\\&\quad \\&={\begin{cases}{\underset {x\to +\infty }{\lim }}{\frac {x{\sqrt {1-{\frac {8}{x^{2}}}}}}{x(1-{\frac {4}{x}})}}={\underset {x\to +\infty }{\lim }}{\frac {\sqrt {1-{\frac {8}{x^{2}}}}}{(1-{\frac {4}{x}})}}=1\\\quad \\{\underset {x\to -\infty }{\lim }}{\frac {-x{\sqrt {1-{\frac {8}{x^{2}}}}}}{x(1-{\frac {4}{x}})}}={\underset {x\to +\infty }{\lim }}{\frac {-{\sqrt {1-{\frac {8}{x^{2}}}}}}{(1-{\frac {4}{x}})}}=-1\end{cases}}\end{aligned}}}
Horizontal asymptotes: y = 1 , y = − 1 {\displaystyle \color {blue}y=1,y=-1}
, the first step is to factor out the highest degree term from numerator and denominator, so we have
