Limits at Infinity and Horizontal Asymptotes
What does a graph do way out at the ends? A limit at infinity answers exactly that — and when the answer is a finite number L, the line y = L is a horizontal asymptote the graph flattens onto. For a rational function, you can read the whole story off two numbers: the degrees of the top and the bottom.
By the end you'll be able to find the end behavior of any rational function by comparing the degrees of its numerator and denominator, and name its horizontal asymptote — or explain why it doesn't have one.
Predict: from just the degrees and leading coefficients below, what's the horizontal asymptote — y = 0, y = a/b, or none at all? Then set the sliders and watch the dashed green line land where you predicted (or vanish).
You're plotting \(f(x) = \dfrac{a\,x^m + 1}{b\,x^n + 1}\) over a wide x-range. The leading coefficients a and b set the a/b ratio the dashed line sits at when the degrees tie. Pick an odd bottom degree and the denominator hits zero at one x-value: if the numerator is nonzero there, a dashed amber line appears — a vertical asymptote, the blow-up-near-a-point kind from the previous lesson, just to keep the two ideas straight. But if the numerator is zero at that same x (try a = b with m = n), the factor cancels and you get a hole with a finite limit instead — no vertical asymptote. Hover or tap the curve for exact values.
Degrees 2 vs 2 (equal) → leading-coefficient ratio 3/6 = 0.50. As x→+∞, f(x) → 0.50; as x→−∞, f(x) → 0.50. Horizontal asymptote y = 0.50.
Limits at infinity describe end behavior and locate horizontal asymptotes; for a rational function the answer comes from comparing the degrees of numerator and denominator.
We write \(\lim_{x \to +\infty} f(x) = L\) or \(\lim_{x \to -\infty} f(x) = L\). If either limit is a finite number L, the line y = L is a horizontal asymptote. A graph may approach the same horizontal asymptote on both ends, different ones on each end, or none at all. The key building block is \(\lim_{x \to \pm\infty} 1/x^n = 0\) for any \(n > 0\) — so the working method for a rational function is to divide top and bottom by the highest power of x and let every \(1/x^n\) term die. What survives depends only on the degrees: with numerator degree m and denominator degree n, m < n (bottom heavier) gives limit 0 and asymptote y = 0; m = n gives the ratio of the leading coefficients; m > n (top heavier) gives \(\pm\infty\) — no horizontal asymptote.
In the explorer above, watch the ends of the curve, not the middle: when a horizontal asymptote exists, both arms flatten onto the dashed green line — they hug it ever closer without needing to touch it. Push the top degree above the bottom and the arms stop flattening and run off the top or bottom of the frame: no line to hug. Don't confuse this with the dashed amber line you saw in the previous lesson — a vertical asymptote is a blow-up near one finite x-value, while a horizontal asymptote is about where the graph settles at the far ends. Two different questions, and one graph can easily have both.
End behavior is the "long run" of a model. Say producing x items costs \(100 + 5x\) dollars, so the average cost per item is \(\frac{100 + 5x}{x}\). Degrees are equal (1 and 1), so as production grows the average cost levels off at the leading-coefficient ratio \(5/1 = 5\) dollars — the fixed $100 gets spread ever thinner. Whenever someone asks "where does this settle in the long run?", they're asking for a limit at infinity, and the horizontal asymptote is the settling level.
Evaluate \(\lim_{x \to \infty} \dfrac{3x^2 - 4x + 1}{6x^2 + x}\). Divide numerator and denominator by the highest power, \(x^2\): \[\lim_{x \to \infty} \frac{3 - 4/x + 1/x^2}{6 + 1/x} = \frac{3 - 0 + 0}{6 + 0} = \frac{1}{2}.\] Degrees are equal (2 and 2), so the answer is the leading-coefficient ratio \(3/6 = 1/2\); the horizontal asymptote is y = 1/2. That's exactly the explorer's starting position — a = 3, b = 6, degrees 2 and 2.
Find the horizontal asymptote of \(f(x) = \dfrac{5x^2 + 1}{2x^2 - 3x}\). The degrees are ____ and ____, so the limit as \(x \to \pm\infty\) is the ratio ____, and the horizontal asymptote is y = ____.
Reveal the answer
Both degrees are 2 — equal — so the limit is the leading-coefficient ratio \(5/2\), and the horizontal asymptote is y = 5/2 = 2.5 on both ends. Check it live: set m = 2, n = 2, a = 5, b = 2 in the explorer and watch the dashed line land at 2.5.
More info — why dividing by the highest power works, and a sign trap at −∞
Dividing top and bottom by \(x^n\) doesn't change the function's value — it just rewrites every term as a constant or a multiple of \(1/x^k\). Since \(\lim_{x \to \pm\infty} 1/x^k = 0\) for \(k > 0\) (the building block above), all the lower-order terms evaporate and only the leading behavior survives. This is the same limit-laws bookkeeping you used earlier in the section, applied at the ends of the axis instead of at a point.
One trap when heading to negative infinity: with even roots, remember \(\sqrt{x^2} = |x|\), and for \(x \to -\infty\) that's \(|x| = -x\), not x — miss the sign and your limit comes out with the wrong sign. Paul's Online Notes (linked in Dive deeper below) works several of these sign-careful examples in full.
Check your understanding
Evaluate \(\lim_{x \to \infty} \dfrac{4x^3 - x}{2x^3 + 5x^2}\).
What is the horizontal asymptote of \(f(x) = \dfrac{2x+5}{x^3-1}\)?
Which statement is true about \(g(x) = \dfrac{x^3 + 1}{x^2 + 4}\) as \(x \to \infty\)?
For \(h(x) = \dfrac{2x+1}{x-3}\), which correctly describes its asymptotes?
Recap
- \(\lim_{x \to \pm\infty} f(x) = L\) (finite) means the graph flattens onto the horizontal asymptote y = L. A graph can approach the same one on both ends, different ones on each end, or none at all.
- Building block: \(\lim_{x \to \pm\infty} 1/x^n = 0\) for \(n > 0\) — divide top and bottom by the highest power of x and the lower-order terms vanish.
- Rational functions, numerator degree m vs denominator degree n: m < n → asymptote y = 0; m = n → y = ratio of leading coefficients; m > n → limit is \(\pm\infty\), no horizontal asymptote.
- Horizontal asymptotes are about the ends (\(x \to \pm\infty\)); vertical asymptotes are blow-ups near a finite x-value — different questions, and one graph can have both.
Dive deeper
- Paul's Online Notes — Limits At Infinity, Part I Compute end-behavior limits and find horizontal asymptotes of rational functions
Sources
- Limits at Infinity and Horizontal Asymptotes