Question

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

Answer 1

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