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

1 Answer
May 20, 2018

Answer:

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

Explanation:

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

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

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

# \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