asymptotes of (x^3-x^2-x+1)/(x^2-4)

1) Let's find the asymptotes of

\frac{x^{3} - x^{2} - x + 1}{x^{2} - 4}

, for real

x

Vertical asymptotes: find the excluded values, where the denominator is zero. Then check whether the curve goes to infinity near these values

⇥1.1) Let's solve in

ℝ

x^{2} - 4 = 0

⇥The two numbers whose square is

4

are

-2

and

2

⇥Solutions obtained:

x = {-2, 2}

x

approaches

-2

from the left (smaller values), the function tends to

-∞

; from the right (larger values), it tends to

+∞

The function tends to infinity on at least one side: the line

x = -2

is therefore a vertical asymptote

x

approaches

2

from the left (smaller values), the function tends to

-∞

; from the right (larger values), it tends to

+∞

The function tends to infinity on at least one side: the line

x = 2

is therefore a vertical asymptote

⇥1.2) Let's start the euclidean division of

(x^{3}

- x^{2}

- x

+ 1)

(x^{2} - 4)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{3}$ $- x^{2}$ $- x$ $+ 1$	$\frac{x^{3}}{x^{2}}$	$x$	$x \cdot (x^{2} - 4) = x^{3} - 4 x$	$(x^{3}$ $- x^{2}$ $- x$ $+ 1) - (x^{3} - 4 x) = - x^{2} + 3 x + 1$
$- x^{2} + 3 x + 1$	$- \frac{x^{2}}{x^{2}}$	$-1$	$-1 \cdot (x^{2} - 4) = - x^{2} + 4$	$(- x^{2} + 3 x + 1) - (- x^{2} + 4) = 3 x - 3$

⇥So the result of the euclidean division is:

(x - 1)

The division lets us write

\frac{x^{3} - x^{2} - x + 1}{x^{2} - 4} = (x - 1) + \frac{3 x - 3}{x^{2} - 4}

: the line to test is

y = x - 1

For a slant asymptote, the difference between the function and the line must tend to 0: check the limit of

\frac{3 x - 3}{x^{2} - 4}

at both infinities

The difference tends to

0

at +∞ and to

0

at -∞: the curve therefore approaches the line at both infinities

The slant asymptote is

y = x - 1

; there is no horizontal asymptote

▶Asymptotes:

{x = -2, x = 2, y = x - 1}

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