Rational Limits at Infinity: Compare Degrees and Leading Coefficients

For a rational function, divide numerator and denominator by the denominator’s highest power of x. Terms containing 1/x, 1/x² and higher reciprocal powers approach zero as |x| grows.

Infinity is a direction, not a number to insert

A limit as x → ∞ describes inputs increasing without bound. It does not mean replacing x by a numerical value called infinity and cancelling symbols. Rewriting the expression in reciprocal powers gives a justified way to identify the dominant terms.

If the numerator’s degree is lower, the limit is zero. If the degrees are equal, the finite limit is the ratio of leading coefficients. If the numerator’s degree is higher, there is no finite horizontal limit in general; determine the direction of divergence or use polynomial division to describe the end behaviour.

Worked example: equal degrees

Find lim as x → ∞ of (6x² − x + 4)/(3x² + 5). Divide every term by x²: the expression becomes (6 − 1/x + 4/x²)/(3 + 5/x²). As x grows, the reciprocal terms approach zero, leaving 6/3 = 2.

The same finite limit occurs as x → −∞ because those reciprocal terms still tend to zero. The graph has horizontal asymptote y = 2 at both ends. This does not mean the function equals 2 everywhere or can never cross that line.

Compare different degrees

For (5x + 1)/(x² + 4), dividing by x² yields (5/x + 1/x²)/(1 + 4/x²), so the limit is zero at either end. For (2x² + 1)/(x + 3), polynomial division gives 2x − 6 + 19/(x + 3). Thus the function tends to +∞ as x → ∞ and −∞ as x → −∞.

The final example has a slant asymptote y = 2x − 6, not a horizontal one. Looking only at the ratio 2/1 would mistakenly report 2 because the powers do not match.

Common mistake

Compare polynomial degrees after considering the whole expression, not simply the first written terms. Keep the sign of x in odd powers when analysing negative infinity. An expression behaving like x changes divergence direction between the two ends, while one behaving like x² does not.

Numerical checks at large positive and negative inputs can support the result, but they do not replace the limiting argument. A function can approach its eventual pattern slowly, especially when coefficients differ greatly.

Degrees same ho tab coefficients ka ratio; warna powers ko bhi dekho.

Three practice questions with worked answers

1. Find lim as x → ∞ of (4x³ + 2)/(2x³ − 7).

   The degrees are equal. Divide by x³; reciprocal terms vanish, leaving 4/2 = 2.

2. Find lim as x → −∞ of (7x + 3)/(x² + 1).

   The numerator degree is smaller. Dividing by x² leaves terms tending to zero over a denominator tending to 1, so the limit is 0.

3. Find lim as x → −∞ of x²/(x + 1).

   Division gives x − 1 + 1/(x + 1). The last term tends to zero while x − 1 tends to −∞, so the limit is −∞.

Related study guides

• Limits and Derivatives Class 11: Standard Limits and Solved Examples
• Divide a Polynomial by a Linear Factor

Need one step explained? Open the DoubtPay study tool. Check the current credit price before requesting a paid answer, and verify the working yourself.

References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Limits at Infinity and Asymptotes. No textbook or exam-board affiliation is claimed.