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

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Answer 1 · 2018-05-20T19:11:11

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

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

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