Unit 2, lesson 31

Horizontal Asymptotes

How the leading terms and a comparison of degrees decide whether a rational function levels off far out, at which height it does so, and why the graph may still cross that line.

  • Theory part 61
  • Practice part 62
  • Rational Functions

Keep pouring salty water into a large tank of fresh water and the mixture gets saltier, yet it never gets saltier than the water you pour in. Its concentration climbs and levels off. A horizontal asymptote is the line that records such leveling off, and for a rational function a quick look at the degrees of the top and the bottom finds it.

Key terms

Horizontal asymptote

Definition Let ff be defined on an interval (c,∞)(c, \infty). We write lim⁡x→∞f(x)=L\lim_{x \to \infty} f(x) = L if for every ε>0\varepsilon > 0 there is a number MM such that

∣f(x)−L∣<εwhenever x>M.\begin{gathered} \lvert f(x) - L \rvert < \varepsilon \\[0.4em] \text{whenever } x > M. \end{gathered}

The limit as x→−∞x \to -\infty is defined the same way on an interval (−∞,c)(-\infty, c), with x<Mx < M in place of x>Mx > M. The line y=Ly = L is a horizontal asymptote of the graph of ff if

lim⁡x→∞f(x)=Lorlim⁡x→−∞f(x)=L.\begin{gathered} \lim_{x \to \infty} f(x) = L \\[0.4em] \text{or} \\[0.4em] \lim_{x \to -\infty} f(x) = L. \end{gathered}

In plain words Far enough to the right, or far enough to the left, the graph stays inside any thin band around the height LL that you choose.

Example A large tank holds 100100 liters of fresh water. Water carrying 3030 grams of salt per liter flows in at 1010 liters per minute, and nothing flows out. After tt minutes the tank holds 300t300t grams of salt in 100+10t100 + 10t liters, so the concentration is

c(t)=300t100+10t=30tt+10 grams per liter.\begin{aligned} c(t) &= \frac{300t}{100 + 10t} \\[0.8em] &= \frac{30t}{t + 10} \text{ grams per liter.} \end{aligned}

Subtracting the level 3030 gives

c(t)−30=−300t+10,c(t) - 30 = -\frac{300}{t + 10},

which tends to 00 as tt grows. So y=30y = 30 is a horizontal asymptote, and since the difference is negative the concentration always stays below 3030. After 9090 minutes it is 2727 grams per liter and after 990990 minutes it is 29.729.7. To stay within 0.10.1 of 3030 the difference needs 300t+10<0.1\frac{300}{t + 10} < 0.1, which holds for every t>2990t > 2990.

Dominant terms

Definition For a polynomial pp of degree nn with leading term anxna_n x^n,

lim⁡x→±∞p(x)anxn=1.\lim_{x \to \pm\infty} \frac{p(x)}{a_n x^n} = 1.

Let r=pqr = \frac{p}{q} be a rational function whose numerator has leading term anxna_n x^n and whose denominator has leading term bmxmb_m x^m. For every x≠0x \ne 0 in the domain,

r(x)=anxnbmxm⋅p(x)/(anxn)q(x)/(bmxm),r(x) = \frac{a_n x^n}{b_m x^m} \cdot \frac{p(x) / (a_n x^n)}{q(x) / (b_m x^m)},

and the second factor tends to 11 as x→∞x \to \infty and as x→−∞x \to -\infty. The leading terms are the dominant terms, and far out r(x)r(x) behaves like anbmxn−m\frac{a_n}{b_m} x^{n - m}, in the sense that the ratio of the two tends to 11. In practice, divide the numerator and the denominator by xmx^m, the highest power of xx in the denominator, and use the fact that cxk→0\frac{c}{x^k} \to 0 as x→±∞x \to \pm\infty for every constant cc and every integer k≥1k \ge 1.

In plain words For large ∣x∣\lvert x \rvert the highest power on top and the highest power on the bottom drown out everything else. Keep those two terms, simplify their ratio, and you know what the function does far out.

Example Take

r(x)=6x2−5x+12x2+7.r(x) = \frac{6x^2 - 5x + 1}{2x^2 + 7}.

The dominant terms give 6x22x2=3\frac{6x^2}{2x^2} = 3. Dividing the top and the bottom by x2x^2 confirms it,

r(x)=6−5x+1x22+7x2→6−0+02+0=3.\begin{aligned} r(x) &= \frac{6 - \frac{5}{x} + \frac{1}{x^2}}{2 + \frac{7}{x^2}} \\[0.8em] &\to \frac{6 - 0 + 0}{2 + 0} = 3. \end{aligned}

The values, rounded to four decimals, close in on 33 rather slowly.

xx101010010010001000
r(x)r(x)2.66182.66182.97402.97402.99752.9975

Degree comparison

Definition Let r=pqr = \frac{p}{q}, where the numerator pp is a nonzero polynomial of degree nn with leading coefficient ana_n and the denominator qq has degree mm and leading coefficient bmb_m.

  1. If n<mn < m, then r(x)→0r(x) \to 0 as x→±∞x \to \pm\infty, and y=0y = 0 is the horizontal asymptote.
  2. If n=mn = m, then r(x)→anbmr(x) \to \frac{a_n}{b_m} as x→±∞x \to \pm\infty, and y=anbmy = \frac{a_n}{b_m} is the horizontal asymptote.
  3. If n>mn > m, then ∣r(x)∣→∞\lvert r(x) \rvert \to \infty as x→±∞x \to \pm\infty, and there is no horizontal asymptote.

Cancelling a common factor lowers both degrees by the same amount and leaves the ratio of the leading coefficients unchanged, so the comparison gives the same answer before and after simplifying.

In plain words Bottom-heavy fractions shrink to 00. Balanced ones level off at the ratio of the leading coefficients. Top-heavy ones grow without bound, so no horizontal line can describe them.

Example Here is one function of each kind.

FunctionDegreesAs x→±∞x \to \pm\infty
3x+1x2+4\frac{3x + 1}{x^2 + 4}1<21 < 2tends to 00
4x3−x2x3+5\frac{4x^3 - x}{2x^3 + 5}3=33 = 3tends to 42=2\frac{4}{2} = 2
x3x2+1\frac{x^3}{x^2 + 1}3>23 > 2no horizontal asymptote

The last function behaves like x3x2=x\frac{x^3}{x^2} = x far out, so it goes to ∞\infty on the right and to −∞-\infty on the left.

Positive and negative directions

Definition The limits as x→∞x \to \infty and as x→−∞x \to -\infty are two separate questions. A function may have a horizontal asymptote in one direction only, or a different one in each direction. Each limit has at most one value, so a graph has at most two horizontal asymptotes. For a rational function the degree comparison shows that if one of the two limits is a finite number LL, the other one is LL as well, so a rational function has at most one horizontal asymptote y=Ly = L. Even then the graph may approach that line from above at one end and from below at the other. The sign of r(x)−Lr(x) - L for large positive xx and for large negative xx decides the side at each end.

In plain words Check the right end and the left end on their own. A rational function that levels off does so at the same height at both ends, but other functions need not.

Example For

f(x)=2x+1x2+1f(x) = \frac{2x + 1}{\sqrt{x^2 + 1}}

use x2=∣x∣\sqrt{x^2} = \lvert x \rvert. For x>0x > 0 dividing the top and the bottom by x=x2x = \sqrt{x^2} gives

f(x)=2+1x1+1x2→2as x→∞.f(x) = \frac{2 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x^2}}} \to 2 \quad \text{as } x \to \infty.

For x<0x < 0 the square root is x2=−x\sqrt{x^2} = -x, and dividing by −x-x gives

f(x)=−2−1x1+1x2→−2as x→−∞.\begin{gathered} f(x) = \frac{-2 - \frac{1}{x}}{\sqrt{1 + \frac{1}{x^2}}} \to -2 \\[0.8em] \text{as } x \to -\infty. \end{gathered}

So the graph has the horizontal asymptote y=2y = 2 on the right and y=−2y = -2 on the left. Indeed f(100)≈2.0099f(100) \approx 2.0099 and f(−100)≈−1.9899f(-100) \approx -1.9899.

Crossing a horizontal asymptote

Definition A horizontal asymptote describes f(x)f(x) only as x→∞x \to \infty or x→−∞x \to -\infty, so nothing stops the graph from meeting the line y=Ly = L at other places, even many times. For a rational function r=pqr = \frac{p}{q} with horizontal asymptote y=Ly = L, the graph meets the line exactly at the points (a,L)(a, L) where aa is in the domain and

p(a)−L q(a)=0.p(a) - L \, q(a) = 0.

Unless rr is constant on its domain, the polynomial p−Lqp - Lq is not the zero polynomial and has degree less than deg⁡q\deg q, so there are at most deg⁡q−1\deg q - 1 such points. Vertical asymptotes are different, since the graph of a rational function never meets one.

In plain words Being an asymptote is about the far ends only. Set the function equal to LL and solve, and every solution in the domain is a place where the graph reaches the line.

Example The function

g(x)=2x2−3xx2+1g(x) = \frac{2x^2 - 3x}{x^2 + 1}

has the horizontal asymptote y=2y = 2, and

g(x)−2=−3x−2x2+1.g(x) - 2 = \frac{-3x - 2}{x^2 + 1}.

This difference is negative for large positive xx and positive for large negative xx, so the graph runs below the line on the right and above it on the left, as g(10)≈1.68g(10) \approx 1.68 and g(−10)≈2.28g(-10) \approx 2.28 show. It switches sides where −3x−2=0-3x - 2 = 0, at x=−23x = -\frac{2}{3}, and indeed

g(−23)=89+249+1=26/913/9=2.g\left(-\frac{2}{3}\right) = \frac{\frac{8}{9} + 2}{\frac{4}{9} + 1} = \frac{26/9}{13/9} = 2.

So the graph crosses its asymptote at the point (−23,2)(-\frac{2}{3}, 2).

Check yourself

  1. Find the horizontal asymptote of the function below.
r(x)=5x2−32−x2r(x) = \frac{5x^2 - 3}{2 - x^2}
Show the answer

Both degrees are 22. In standard form the denominator is −x2+2-x^2 + 2, so its leading coefficient is −1-1, not 22. The horizontal asymptote is y=5−1=−5y = \frac{5}{-1} = -5 in both directions.

  1. Does the function below have a horizontal asymptote? What does it do far out?
r(x)=x3+1x2−4r(x) = \frac{x^3 + 1}{x^2 - 4}
Show the answer

No. The numerator has degree 33 and the denominator degree 22, so for large ∣x∣\lvert x \rvert the function behaves like x3x2=x\frac{x^3}{x^2} = x. Thus r(x)→∞r(x) \to \infty as x→∞x \to \infty and r(x)→−∞r(x) \to -\infty as x→−∞x \to -\infty.

  1. Where does the graph of the function below cross its horizontal asymptote?
r(x)=x2−4xx2−x−2r(x) = \frac{x^2 - 4x}{x^2 - x - 2}
Show the answer

The degrees are equal and both leading coefficients are 11, so the asymptote is y=1y = 1. Setting x2−4x=x2−x−2x^2 - 4x = x^2 - x - 2 gives −3x=−2-3x = -2, so x=23x = \frac{2}{3}. The denominator (x−2)(x+1)(x - 2)(x + 1) is not zero there, so the graph crosses the asymptote at (23,1)(\frac{2}{3}, 1).

How this lesson connects

Where this leads in Calculus