asymptotes of f(x)=((x^2+1))/(2x^2+7)

1) Let's find the asymptotes of

\frac{x^{2} + 1}{2 x^{2} + 7}

, 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} + 7 = 0

⇥ⓘ Both sides are divided by

2

⇥

x^{2} = - \frac{7}{2}

⇥In ℝ, a square is never negative: no real number has

- \frac{7}{2}

as its square

⇥Solution:

x \in \emptyset

⇥1.2) Let's start the euclidean division of

(x^{2} + 1)

(2 x^{2} + 7)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{2} + 1$	$\frac{x^{2}}{2 x^{2}}$	$\frac{1}{2}$	$\frac{1}{2} \cdot (2 x^{2} + 7) = x^{2} + \frac{7}{2}$	$(x^{2} + 1) - (x^{2} + \frac{7}{2}) = - \frac{5}{2}$

⇥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} + 1}{2 x^{2} + 7}

at both infinities

x

tends to +∞,

\frac{x^{2} + 1}{2 x^{2} + 7}

tends to

\frac{1}{2}

; as

x

tends to -∞,

\frac{x^{2} + 1}{2 x^{2} + 7}

tends to

\frac{1}{2}

The horizontal asymptote is

y = \frac{1}{2}

; there is no slant asymptote

▶Asymptotes:

y = \frac{1}{2}

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