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

1) Let's determine the asymptotes of

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

, 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) We solve in

ℝ

x - 1 = 0

⇥ⓘ Add

1

to both sides

⇥Solution found:

x = 1

x

approaches

1

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 = 1

is therefore a vertical asymptote

⇥1.2) Let's start the Euclidean division of

(x^{2} + 1)

(x - 1)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{2} + 1$	$\frac{x^{2}}{x}$	$x$	$x \cdot (x - 1) = x^{2} - x$	$(x^{2} + 1) - (x^{2} - x) = x + 1$
$x + 1$	$\frac{x}{x}$	$1$	$x - 1 = x - 1$	$(x + 1) - (x - 1) = 2$

⇥Therefore, the result of the Euclidean division is:

(x + 1)

The division lets us write

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

: 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{2}{x - 1}

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

; therefore, there is no horizontal asymptote

▶Asymptotes:

{x = 1, y = x + 1}

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