# Horizontal Asymptotes by rogerholland

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```									Horizontal Asymptotes
Keerthi Chandrasekaran

Horizontal asymptotes are results of the following limit (for rational functions)
x→+∞

lim f (x) = a or lim f (x) = b
x→−∞ h(x) . g(x)

TThe trick idenify a HA of f (x) =

If h(x) and g(x) is of same degree then the function has a HA, if the degree on the denominator g(x) higher than that of numerator h(x) the horizontal asymptote is y = 0 (the x-axis). The formal solution would depend on the following limit identities. 1 = 0 n = 0, 1, 2, 3.. x→±∞ xn lim Example(1):
x→+∞

lim f (x) =

2x2 + 3 x→+∞ x2 + 7 lim
2x2 +3 x2 lim x2 +7 x→+∞ x2 3 x2 7 x2

=

2+ = lim x→+∞ 1 + = = 1
2+3×0 1+7×0 2 7

The same applies for x → −∞. HA for this problem would be y = 2/7

Example (2): In the next example we get two diﬀerent HA.
x→+∞

lim f (x) =

x→+∞

lim √

x−1 4x2 + 1
1 x(1 − x )

=

x→+∞

lim

x2 (4 +

1 ) x2

=

x→+∞

1 |x| (4 + x2 ) √ because x2 = |x|

lim

1 x(1 − x )

on x → +∞ |x| = x when x > 0 Need this step for full credit 1 x(1 − x ) = lim x→+∞ 1 (+x) (4 + x2 )
1 (1 − x )

=

x→+∞

lim

(4 +

1 ) x2

(1 − 0) = √ 4+0 = so y = 1/2 is a HA. 1 2

2

Now in case of x → −∞ we get
x→−∞

lim f (x) =

x→−∞

lim √

x−1 4x2 + 1
1 x(1 − x )

=

x→−∞

lim

x2 (4 +

1 ) x2

=

x→−∞

1 |x| (4 + x2 ) √ because x2 = |x|

lim

1 x(1 − x )

on x → −∞ |x| = −x when x < 0 Need this step for full credit 1 x(1 − x ) = lim x→−∞ 1 (−x) (4 + x2 )
1 −(1 − x )

=

x→−∞

lim

(4 +

1 ) x2

=

−(1 − 0) √ 4+0 −1 2

= HA :y = +1/2, y = −1/2

3

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