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A limit as x → ∞ asks what happens to f(x) when x gets larger than any number you name. For a ratio of polynomials, the answer is decided almost entirely by the highest powers. Students who know this get these questions in one line; students who do not try to substitute "∞" and end up with ∞/∞. This post gives the rule, explains why it works, and covers the square-root cases that trip people up when x → −∞.
Is this in NCERT and the JEE syllabus?
The 2026-27 NCERT Class 11 chapter Limits and Derivatives treats limits of polynomial and rational functions as x approaches a number a. It mentions a limit equal to +∞ once, and then says "we will not come across such limits in this course." Limits as x → ∞ are not treated. The Syllabus for JEE (Main) 2026 lists "Limits, continuity and differentiability" in Unit 7 without restricting the kind of limit, and the JEE (Advanced) 2026 brochure lists the limit of a function at a real number and L'Hospital's rule. So the method here belongs to the general limits unit for JEE Main, and it also helps with ∞/∞ forms and with drawing graphs. Check your own exam year's documents.
What is the rule?
Write f(x) = p(x)/q(x), where p has degree m and leading coefficient a, and q has degree n and leading coefficient b. As x → ∞:
m < n: the limit is 0.
m = n: the limit is a/b.
m > n: f(x) grows without bound, to +∞ or −∞ depending on the signs of a/b and of x.
Why it works. Divide every term, top and bottom, by xⁿ, the highest power in the denominator. Every term with a negative power of x tends to 0, so only the leading terms survive.
Worked examples
1. Equal degrees. lim (3x² − 5x + 1)/(6x² + 7) as x → ∞. Divide by x²: (3 − 5/x + 1/x²)/(6 + 7/x²) → 3/6 = 1/2.
2. Bottom wins. lim (4x³ + 1)/(2x⁵ − x) as x → ∞. Divide by x⁵: (4/x² + 1/x⁵)/(2 − 1/x⁴) → 0/2 = 0.
3. Hidden leading terms. lim (2x + 1)(x − 3)/(5x²) as x → ∞. Do not expand everything. The leading term on top is 2x · x = 2x², so the limit is 2/5.
4. Top wins, and what the graph does. f(x) = (x³ + 2)/(x² + 1). The degree on top is higher, so f(x) → ∞ as x → ∞ and f(x) → −∞ as x → −∞. Long division gives more: x³ + 2 = x(x² + 1) − x + 2, so f(x) = x + (2 − x)/(x² + 1). The fraction tends to 0, so the graph hugs the line y = x for large |x|. That line is an oblique asymptote.
5. A sum that becomes a rational function. lim (1² + 2² + … + n²)/n³ as n → ∞. The sum of squares is n(n + 1)(2n + 1)/6, a cubic with leading term 2n³/6. So the limit is (2/6)/1 = 1/3.
Why does x → −∞ change the answer with square roots?
For rational functions, the answers 0 and a/b are the same as x → ∞ and as x → −∞. Square roots change that, because √(x²) = |x|, not x.
6. Find lim (√(x² + 1) + 3x)/(2x − 5) as x → ∞ and as x → −∞.
For x > 0, √(x² + 1) = x√(1 + 1/x²). Divide top and bottom by x: (√(1 + 1/x²) + 3)/(2 − 5/x) → (1 + 3)/2 = 2.
For x < 0, √(x² + 1) = |x|√(1 + 1/x²) = −x√(1 + 1/x²). Divide by x: (−√(1 + 1/x²) + 3)/(2 − 5/x) → (−1 + 3)/2 = 1.
So the graph has two different horizontal asymptotes, y = 2 on the right and y = 1 on the left. We checked both values numerically at x = ±10⁸.
Which mistakes cost the most marks?
Writing ∞/∞ = 1. It is an indeterminate form. Divide by the highest power first.
Dividing top and bottom by different powers. Divide both by the same power, usually the highest power in the denominator.
Forgetting |x| at −∞. Any time a square root of x² appears, ask which sign x has.
Expanding products you do not need. Multiply only the leading terms.
Reporting "∞" without a sign. In example 4, the two directions go to different infinities.
For the other indeterminate forms and the standard limits, see our post on limits and continuity for JEE.
A practice routine
Studying worked examples before solving can speed up learning a new procedure. Sweller and Cooper (1985) found that algebra students who studied worked examples then solved similar problems faster and with fewer errors, although the benefit was limited to problems with the same structure. That last point is why the mixed set below matters.
Day 1: Study examples 1 to 3 above, then write the rule from memory.
Day 2: Ten limits at +∞ with random degrees. Before working, predict each answer from degrees alone.
Day 3: Five square-root limits, each at both +∞ and −∞.
Day 4: A mixed set of limits at a point and at infinity, so you have to choose the method.
If you keep getting stuck at the set-up, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can see why a particular power was divided out.
Before your next maths session, find the limit of (5x⁴ − x)/(1 − 2x⁴) as x → ∞, and check that you get −5/2.
