How do you find vertical, horizontal and oblique asymptotes for [e^(x)-2x] / [7x+1]ex2x7x+1?

1 Answer
May 20, 2018

Vertical Asymptote: x = \frac{-1}{7} x=17
Horizontal Asymptote: y = \frac{-2}{7} y=27

Explanation:

Vertical Asymptotes occur when the denominator gets extremely close to 0:

Solve 7x+1 = 0, 7x = -1 7x+1=0,7x=1

Thus, the vertical asymptote is
x = \frac{-1}{7} x=17

\lim _{x\to +\infty }(\frac{e^x-2x}{7x+1}) = e^x
No Asymptote

\lim _{x\to -\infty }(\frac{e^x-2x}{7x+1}) = \lim _{x\to -\infty }\frac{0-2x}{7x} = \frac{-2}{7}

Thus there is a horizontal aysmptote at y = \frac{-2}{7}

since there is a horizontal aysmptote, there are no oblique aysmptotes