asymptotes of f(x)=(x^2)/(2x^2-2)

⚠Ⓘ 2x is read as the product 2*x: for a power, write 2^x

1) Let's find the asymptotes of

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

, 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

2 x^{2} - 2 = 0

⇥You can write the expression in factored form:

2 \cdot (x^{2} - 1) = 0

⇥1.2) Let's solve:

x^{2} - 1 = 0

⇥Solutions found:

x = {-1, 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

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

⇥1.3) Compute the quotient and remainder of

x^{2}

2 x^{2} - 2

⇥

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

⇥So the result of the euclidean division is:

\frac{1}{2}

The quotient of the division is the constant

\frac{1}{2}

: the line to test is

y = \frac{1}{2}

For a horizontal asymptote, the function must tend to a finite value: check the limit of

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

at both infinities

x

tends to +∞,

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

tends to

\frac{1}{2}

; as

x

tends to -∞,

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

tends to

\frac{1}{2}

The horizontal asymptote is

y = \frac{1}{2}

; there is no slant asymptote

▶Asymptotes:

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

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