Limits: Solved Examples for JEE Main Maths

Limits is the foundation of calculus in JEE Main mathematics, contributing 1–2 direct questions per session and underpinning continuity, differentiation, and integration. JEE Main tests limits through indeterminate forms (0/0, infinity/infinity, 0·infinity, 1^infinity, 0^0, infinity^0), algebraic and trigonometric limits, limits involving exponentials and logarithms, and sandwich theorem applications. This guide provides 40 solved examples across all major limit types, with the efficient methods that reduce a 5-minute problem to a 90-second solution.

Algebraic Limits and Standard Substitution

For limits where direct substitution works (no 0/0 form): simply substitute x = a and evaluate. Example: lim(x→2)(3x² − 5x + 1) = 3(4) − 5(2) + 1 = 3. For rational functions where substitution gives 0/0: factorise and cancel. Example: lim(x→3)(x²−9)/(x−3) = lim(x→3)(x+3)(x−3)/(x−3) = lim(x→3)(x+3) = 6. Standard algebraic limit: lim(x→a)(x^n − a^n)/(x−a) = n·a^(n−1) (valid for all real n). Example: lim(x→2)(x^5 − 32)/(x−2) = 5·2^4 = 80. For the continuity and differentiability chapter that builds directly on limit concepts, see our Continuity and Differentiability Guide.

Rationalization for limits involving surds: multiply numerator and denominator by the conjugate. Example: lim(x→0)(sqrt(1+x) − 1)/x. Multiply by (sqrt(1+x)+1)/(sqrt(1+x)+1): = lim(x→0)(1+x−1)/(x(sqrt(1+x)+1)) = lim(x→0)1/(sqrt(1+x)+1) = 1/2. Alternatively, use the expansion (1+x)^(1/2) approximately 1 + x/2 for small x, giving (1 + x/2 − 1)/x = 1/2. The expansion method (Binomial series) is generally faster than rationalization for JEE Main and worth mastering.

Trigonometric Limits and Standard Results

Standard trigonometric limits (all as theta → 0): lim(sin(theta)/theta) = 1; lim(tan(theta)/theta) = 1; lim((1−cos(theta))/theta²) = 1/2; lim(sin(n·theta)/(n·theta)) = 1 for any constant n. These four results (and their combinations) solve 80% of JEE Main trigonometric limit problems. Example: lim(x→0)(sin(3x))/(5x). Rewrite as (sin(3x))/(3x) × 3/5. As x→0, sin(3x)/(3x) → 1. So limit = 3/5. Example: lim(x→0)(1−cos(2x))/x². Using (1−cos(2x))/x² = 2sin²(x)/x² = 2(sin(x)/x)² → 2·1² = 2. Practise trigonometric and algebraic limits on our JEE Main mathematics mock tests with instant answer checking and solution review.

For limits involving tan, cot, sec, cosec as x → pi/2 or x → 0: (1) lim(x→0)(x·csc(x)) = lim(x→0)(x/sin(x)) = 1. (2) lim(x→pi/2)(pi/2 − x)·tan(x) = lim(t→0)(t/tan(pi/2−t)) = lim(t→0)(t·tan(t)/t) ... use substitution t = pi/2−x. These require substitution first, then standard results. Practice 10 such problems with substitution until the approach is automatic.

Exponential, Logarithmic, and 1^infinity Forms

Standard exponential limits: lim(x→0)(e^x−1)/x = 1; lim(x→0)(a^x−1)/x = ln(a); lim(x→0)(ln(1+x))/x = 1. Derived results: lim(x→0)(a^x−b^x)/x = ln(a/b). These limits are fundamental and appear directly in JEE Main. Example: lim(x→0)(3^x − 2^x)/x = ln(3/2). The 1^infinity form: lim(x→0)(1+x)^(1/x) = e. General form: lim f(x)^(g(x)) where f→1 and g→infinity — rewrite as exp(lim(g(x)·(f(x)−1))). If the exponent limit is L, the original limit is e^L. Example: lim(x→0)(1+2x)^(3/x) = e^(lim(x→0)(3/x)·2x) = e^6. Example: lim(x→0)(cos(x))^(1/x²). Since cos(x) → 1 and 1/x² → infinity: exponent limit = lim(1/x²)·(cos(x)−1) = lim(1/x²)·(−x²/2) = −1/2. So limit = e^(−1/2) = 1/sqrt(e).

L'Hopital's rule: for 0/0 or infinity/infinity forms, lim(f(x)/g(x)) = lim(f'(x)/g'(x)), provided the latter limit exists. Apply repeatedly if needed. Important: L'Hopital requires differentiability near the limit point, and both f and g must tend to 0 (or both to infinity). Do not apply L'Hopital to forms like 0/1 or 1/infinity — these are not indeterminate. Example: lim(x→0)(e^x − x − 1)/x². Using L'Hopital twice: first application gives (e^x − 1)/(2x); second application gives e^x/2 → 1/2. Alternatively, use expansion: e^x ≈ 1 + x + x²/2 + ..., so e^x − x − 1 ≈ x²/2, and (e^x−x−1)/x² → 1/2. The expansion method is often faster and less error-prone than repeated L'Hopital.

Expansion Methods and Sandwich Theorem

Taylor/Maclaurin expansions essential for JEE Main limits: e^x = 1 + x + x²/2! + x³/3! + ...; sin(x) = x − x³/3! + x⁵/5! − ...; cos(x) = 1 − x²/2! + x⁴/4! − ...; ln(1+x) = x − x²/2 + x³/3 − ...; (1+x)^n = 1 + nx + n(n−1)x²/2! + ... (Binomial). For limits as x → 0, keep only the lowest-order term in the numerator and denominator. Example: lim(x→0)(sin(x) − x)/x³. sin(x) − x ≈ −x³/6, so limit = −1/6. This expansion approach solves the problem in 20 seconds vs. 3 applications of L'Hopital. Sandwich (Squeeze) Theorem: if g(x) ≤ f(x) ≤ h(x) for x near a, and lim(g(x)) = lim(h(x)) = L, then lim(f(x)) = L. Classic example: lim(x→0)(x·sin(1/x)) — since −|x| ≤ x·sin(1/x) ≤ |x| and |x| → 0, the limit is 0. Register on our platform to access the complete limits question bank with 200+ problems. Our premium subscription covers all JEE Main calculus chapters. For the differentiability chapter that directly extends limit concepts, see our Continuity and Differentiability Guide.

Exam tip: for JEE Main limits problems, try expansion method first (10 seconds), then substitution/factorisation, then L'Hopital (as a last resort). The expansion method is the fastest approach for 70% of JEE Main limits and requires only memorising 5 standard expansions (e^x, sin, cos, ln(1+x), binomial). Invest 1 hour memorising and practising these expansions and save 5–8 minutes on the calculus section of every mock test.