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

May 20, 2018

Vertical Asymptote: $x = \setminus \frac{- 1}{7}$
Horizontal Asymptote: $y = \setminus \frac{- 2}{7}$

#### Explanation:

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

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

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

$\setminus {\lim}_{x \setminus \to + \setminus \infty} \left(\setminus \frac{{e}^{x} - 2 x}{7 x + 1}\right) = {e}^{x}$
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 $y = \setminus \frac{- 2}{7}$

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