Limits at Infinity and Horizontal Asymptotes

You have a rational function, and you need to know where it goes at the far ends of the x-axis. That is a limit at infinity problem: as x heads toward positive or negative infinity, does the function settle on a horizontal line, blow up, or vanish? The answer tells you the function's end behavior and, if the function levels off, gives you its horizontal asymptote.

The core tool is a three-way comparison of degrees, the highest exponents in the numerator and denominator. The three degree rules for rational functions cover every case, and they work for both +∞ and −∞ once you handle the sign correctly. The algebraic method of dividing by the highest power explains those rules, extends the logic to radicals and exponentials, and connects the result directly to horizontal asymptotes.

Let n be the degree of the numerator and m the degree of the denominator. These three rules determine the limit as x → ±∞ for any rational function (ratio of polynomials):

• n < m: The limit is 0. The denominator grows faster, so the fraction shrinks to zero. Example: lim_{x→∞} (x)/(x² + 1) = 0.
• n = m: The limit equals the ratio of the leading coefficients. Example: lim_{x→∞} (2x²)/(3x²) = 2/3.
• n > m: The limit is infinite (∞ or −∞). The numerator dominates. Example: lim_{x→∞} (x³)/(x² + 1) = ∞.

The first two rules produce a finite number, so they give you a horizontal asymptote. The third rule produces an infinite limit, meaning the function does not have a horizontal asymptote; it grows or decays without bound.

Worked Example: n < m

Find lim_{x→∞} (4x + 2)/(x³ − x). The numerator degree is 1, denominator degree is 3. Since 1 < 3, the limit is 0. No horizontal asymptote? Wait, there is one: y = 0 is the horizontal asymptote because the function approaches 0 as x gets large.

Worked Example: n = m

Find lim_{x→∞} (3x² + 5x)/(2x² − 7). Degrees equal (both 2). Leading coefficients: numerator = 3, denominator = 2. Limit = 3/2. The horizontal asymptote is y = 3/2.

Worked Example: n > m

Find lim_{x→∞} (x⁴ − 3x)/(2x² + 1). Numerator degree 4 > denominator degree 2. The limit is ∞. No horizontal asymptote exists because the function grows without bound.

The algebraic method that justifies the degree rules is dividing by the highest power. You divide every term in the numerator and denominator by x^m, where m is the denominator's degree. This transforms the function into one where the lower-degree terms vanish as x → ∞, leaving only the leading behavior.

Example: lim_{x→∞} (2x² + 3x − 1)/(x² − 4). Divide numerator and denominator by x²: (2 + 3/x − 1/x²)/(1 − 4/x²). As x → ∞, 3/x → 0, 1/x² → 0, 4/x² → 0, leaving 2/1 = 2. The horizontal asymptote is y = 2.

Why This Works

At infinity, only the highest-degree term in each polynomial matters. Lower-degree terms are negligible by comparison. Dividing by the highest power isolates that term and lets the rest drop out. This method also works when the degrees are different, confirming the three rules.

A horizontal asymptote limit is the finite value a function approaches as x → ∞ or x → −∞. It is the y-value of the horizontal line the function gets arbitrarily close to. For rational functions, this limit exists exactly when the degree of the numerator is less than or equal to the degree of the denominator.

If the limit is L, then y = L is the horizontal asymptote. A function can have two different horizontal asymptotes, one for +∞ and one for −∞, though for rational functions they are usually the same unless signs cause a split. For example, f(x) = (x)/(x² + 1) has y = 0 as its only horizontal asymptote for both directions.

When you evaluate a limit as x approaches infinity, you are asking what happens to f(x) as x grows without bound in the positive direction. The same question applies for negative infinity. The answer can be a finite number, ∞, −∞, or the limit may not exist (DNE) due to oscillation.

For rational functions, the limit as x → ∞ is determined entirely by the leading terms. Polynomials themselves are a special case: the limit of a polynomial at infinity is determined by its leading term. For example, lim_{x→∞} (x³ − 5x) = ∞ because the leading term x³ dominates.

Sign Trap at Negative Infinity

When x → −∞, the sign of odd-powered terms flips. For a polynomial with an odd-degree leading term, the sign at −∞ is opposite to the sign at +∞. For rational functions, check the sign of the leading term's coefficient and the parity of the degree. Example: lim_{x→−∞} (x³)/(x²) = lim_{x→−∞} x = −∞, not +∞.

End behavior limits are the general term for limits as x → ±∞. They describe what the function does at the extreme ends of its domain. For rational functions, end behavior limits are the same as the limits at infinity, they give you the horizontal asymptote or the unbounded growth.

But end behavior limits also apply to non-rational functions. Radicals and exponentials require their own handling because they grow at different rates than polynomials. The growth-order table below ranks common function types by how fast they increase as x → ∞.

Speed	Type	Example
Fastest	Exponential (base >1)	e^x, 2^x, 10^x
Fast	Power polynomial (high degree)	x^n for n large
Medium	Power polynomial (low degree)	x, x²
Slow	Logarithmic	ln(x), log₂(x)
Slowest	Constant or decaying	1/x, 1/x²

When a rational function contains a radical, you still divide by the highest power of x, but you must handle the square root carefully. For √(x²), the result is |x|, not x. This matters at negative infinity because √(x²) = −x when x is negative.

Example: lim_{x→−∞} (√(x² + 1))/(x). Divide numerator and denominator by x: (√(x² + 1))/x = (√(x²(1 + 1/x²)))/x = (|x|√(1 + 1/x²))/x. Since x → −∞, |x| = −x, so the expression becomes (−x√(1 + 1/x²))/x = −√(1 + 1/x²). As x → −∞, this approaches −1. The limit is −1, and y = −1 is the horizontal asymptote.

If you had treated √(x²) as x, you would have gotten +1, which is wrong. Always rewrite √(x²) as |x| when x might be negative. The same principle applies to any even root.

Exponential functions like e^x or 2^x grow faster than any polynomial as x → ∞. This means for a rational function that mixes exponentials and polynomials, the exponential term dominates completely. To find the limit, factor out the dominant exponential term from numerator and denominator.

Example: lim_{x→∞} (e^x + x)/(2e^x − 1). Factor e^x from numerator and denominator: e^x(1 + x/e^x)/[e^x(2 − 1/e^x)] = (1 + x/e^x)/(2 − 1/e^x). As x → ∞, x/e^x → 0 and 1/e^x → 0, so the limit is 1/2.

Logarithmic functions grow much slower than any positive power of x. For example, ln(x)/x → 0 as x → ∞. This makes logarithms negligible compared to polynomials.

Comparison to Polynomials

The growth order from fastest to slowest: exponentials (base > 1) > positive powers of x (xⁿ) > logarithms (ln x). This ranking determines end behavior when functions from different families are combined.

The limits at infinity you compute are the same numbers that define horizontal asymptotes. If lim_{x→∞} f(x) = L, then the line y = L is a horizontal asymptote. If the limit is ∞ or −∞, there is no horizontal asymptote.

A function can have at most two horizontal asymptotes, one for the right end and one for the left end. For rational functions, they are often the same. For functions with radicals or exponentials, they may differ.

Common error: a horizontal asymptote is not the same as a limit. The limit is the value the function approaches; the asymptote is the line y = that value. The function can cross the asymptote, though for rational functions it usually does not after a certain point.