Limits at Infinity in Calculus: A Comprehensive Guide with Examples, Study notes of Calculus for Engineers

Similarly, a limit equals −∞ when f(x) becomes lower than −B for any large B. At the end of §3.3, we saw how a sign chart for f (x) can classify vertical.

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Math 132 Limits at Infinity Stewart §3.4
Vertical asymptotes. We say a curve has a line as an asymptote if, as the curve
runs outward to infinity, it gets closer and closer to the line. “Closer and closer”
reminds us of limits, and indeed we have seen that x=ais a vertical asymptote
of y=f(x) whenever one of the following holds:
lim
xa
f(x) = lim
xa+
f(x) = lim
xa
f(x) = −∞ lim
xa+
f(x) = −∞.
As we saw in §1.5, has no meaning by itself; rather, the whole equation means
that, as xgets closer to (but unequal to) a, the output f(x) eventually becomes
higher than any given bound B, such as B= 100 or 1000 or 1 billion. Similarly,
a limit equals −∞ when f(x) becomes lower than Bfor any large B.
At the end of §3.3, we saw how a sign chart for f0(x) can classify vertical
asymptotes. We could do this with a sign chart for f(x) itself, with no derivatives.
example: Let:
f(x) = x26x+9
x36x2+11x6=(x3)2
(x1)(x2)(x3) =x3
(x1)(x2) .
(To determine vertical asymptotes and intercepts, we always want f(x) in factored
form.) In the original form, the denominator vanishes at x= 3, but we work with
the cancelled form at right.
The function can only change its sign at points where f(x) = 0 (numerator =
0) or f(x) is not defined (denominator = 0), that is, x= 1,2,3. In the interval
x(−∞,1), the sign is given by a sample point like f(0) = 2
(1)(3) =2
3<0,
so f(x) is negative; and similarly for the other intervals.
x1 2 3
f(x) ±∞ +±∞ 0 +
Each time xpasses one of the sign-change candidates x=a, a factor (xa) changes
from negative to positive, and f(x) does indeed change sign.
Notes by Peter Magyar [email protected]
To factor the bottom, we try linear factors xm
n, where mis an integer fac-
tor of the constant coefficient 6, and nis an integer factor of the highest coefficient
1, so n=±1,±2,±3,±6 and m=±1. Trying m
n= 1, we find x1 is a fac-
tor, since polynomial long division gives x36x2+11x6=(x1)(x25x+6), and the
quadratic is easy to factor. For a review of polynomial long division, see Khan Academy:
www.khanacademy.org/math/algebra2/polynomial and rational/dividing polynomials/v/polynomial-division.
pf3
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Math 132 Limits at Infinity Stewart §3.

Vertical asymptotes. We say a curve has a line as an asymptote if, as the curve runs outward to infinity, it gets closer and closer to the line. “Closer and closer” reminds us of limits, and indeed we have seen that x = a is a vertical asymptote of y = f (x) whenever one of the following holds:

lim x→a−

f (x) = ∞ lim x→a+

f (x) = ∞ lim x→a−

f (x) = −∞ lim x→a+

f (x) = −∞.

As we saw in §1.5, ∞ has no meaning by itself; rather, the whole equation means that, as x gets closer to (but unequal to) a, the output f (x) eventually becomes higher than any given bound B, such as B = 100 or 1000 or 1 billion. Similarly, a limit equals −∞ when f (x) becomes lower than −B for any large B. At the end of §3.3, we saw how a sign chart for f ′(x) can classify vertical asymptotes. We could do this with a sign chart for f (x) itself, with no derivatives.

example: Let:

f (x) = x^2 − 6 x+ x^3 − 6 x^2 +11x− 6

(x−3)^2 (x−1)(x−2)(x−3)

x− 3 (x−1)(x−2)

(To determine vertical asymptotes and intercepts, we always want f (x) in factored∗ form.) In the original form, the denominator vanishes at x = 3, but we work with the cancelled form at right. The function can only change its sign at points where f (x) = 0 (numerator =

  1. or f (x) is not defined (denominator = 0), that is, x = 1, 2 , 3. In the interval x ∈ (−∞, 1), the sign is given by a sample point like f (0) = (^) (−1)(−^2 −3) = − 23 < 0, so f (x) is negative; and similarly for the other intervals.

x 1 2 3 f (x) − ±∞ + ±∞ − 0 +

Each time x passes one of the sign-change candidates x = a, a factor (x−a) changes from negative to positive, and f (x) does indeed change sign.

Notes by Peter Magyar [email protected] ∗ (^) To factor the bottom, we try linear factors x − m n ,^ where^ m^ is^ an^ integer^ fac- tor of the constant coefficient 6, and n is an integer factor of the highest coefficient 1, so n = ± 1 , ± 2 , ± 3 , ± 6 and m = ±1. Trying mn = 1, we find x− 1 is a fac- tor, since polynomial long division gives x^3 − 6 x^2 +11x− 6 = (x−1)(x^2 − 5 x+6), and the quadratic is easy to factor. For a review of polynomial long division, see Khan Academy: www.khanacademy.org/math/algebra2/polynomial and rational/dividing polynomials/v/polynomial-division.

Here f (x) = ±∞ just means the denominator vanishes and there is a vertical asymptote. The signs on each side of the asymptote show whether the graph shoots upward or downward: we have limx→ 1 − f (x) = −∞, limx→ 1 + f (x) = ∞, limx→ 2 − f (x) = ∞, limx→ 2 + f (x) = −∞.

Horzontal asymptotes. To understand the behavior of the graph over the left and right ends of the x-axis, we will need a new kind of limit in which x becomes larger and larger.

Definition:

  • limx→∞ f (x) = L means that f (x) can be forced arbitrarily close to L, closer than any given ε > 0, by making x > B for some B.
  • limx→−∞ f (x) = L means that f (x) can be forced arbitrarily close to L, closer than any given ε > 0, by making x < −B for some B.

Graphically, limx→∞ f (x) = L means that toward the right of the x-axis, the graph y = f (x) approaches the horizontal asymptote y = L; and similarly for limx→−∞ f (x) = L toward the left. We can even have limx→∞ f (x) = ∞, which means that the graph goes off toward the upper right of the xy-plane in an un- specified way. The most basic x → ∞ limits are the power funcitons: for a positive real number power p > 0, we have:†

lim x→∞ xp^ = ∞, lim x→∞

xp^

For x → −∞, consider the rational power p = mn where m, n are positive integers with n odd (perhaps n = 1); then:

lim x→−∞ xm/n^ =

∞ for m even −∞ for m odd, lim x→−∞

xm/n^

† (^) Proof: For any large bound C, we can force xp (^) > C if we take x so large that x > C 1 /p. For any small error tolerance ε > 0, we can force | (^) x^1 p − 0 | < ε if we take x so large that x > ( (^1) ε )^1 /p.

example: For

f (x) = x^2 + 3x^7 /^2 − x−^5 9 x

x + 4x^2

x

the terms in the denominator are 9xx^1 /^2 = 9x^3 /^2 and 4x^2 x^1 /^2 = 4x^5 /^2 , so the second is the highest term. Thus:

xlim→∞ f^ (x)^ =^ xlim→∞

x^2 + 3x^7 /^2 − x−^5 9 x

x + 4x^2

x = (^) xlim→∞ 3 x^7 /^2 4 x^5 /^2

= lim x→∞

x^7 /^2 −^5 /^2 = lim x→∞

x = ∞,

which means y = f (x) has no horizontal asymptote. However, the approximation f (x) ≈ 34 x implies that the right end of the graph looks like a line with slope 34. (See slant asymptotes in §3.5.) This function is not defined for x < 0, so there is no left end.