Limits

🟢 Lite — Quick Review

Rapid summary for last-minute revision before your exam.

Limits — Key Facts for JEE Main Limit: value that f(x) approaches as x approaches a (may or may not equal f(a)) Limit exists: left-hand limit = right-hand limit (LHL = RHL) Indeterminate forms: 0/0, ∞/∞, 0·∞, ∞−∞, 0⁰, ∞⁰, 1^∞ Standard limits: lim(x→0) sin x/x = 1; lim(x→0) (e^x − 1)/x = 1; lim(x→0) (a^x − 1)/x = ln a ⚡ Exam tip: L’Hôpital’s rule is very useful for 0/0 and ∞/∞ forms — differentiate numerator and denominator separately!

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🟡 Standard — Core Study

Standard content for students with a few days to months.

Limits — JEE Main Study Guide

Standard limit formulas:

• lim(x→0) sin x/x = 1
• lim(x→0) (tan x)/x = 1
• lim(x→0) (e^x − 1)/x = 1
• lim(x→0) ln(1+x)/x = 1
• lim(x→0) (a^x − 1)/x = ln a
• lim(x→0) (1 + x)^(1/x) = e
• lim(x→∞) (1 + 1/x)^x = e

Algebraic manipulation:

• Factorise: if limit is 0/0, try factorising numerator and denominator
• Rationalise: for square root expressions, multiply by conjugate
• Divide by highest power: for ∞/∞ form, divide numerator and denominator by highest power of x

L’Hôpital’s Rule: For 0/0 or ∞/∞: lim f(x)/g(x) = lim f’(x)/g’(x) if limit exists Can apply repeatedly if needed Only valid when limit is of indeterminate form!

Exponential limits:

• lim(x→0) (1 + x)^(1/x) = e
• lim(x→∞) (1 + a/x)^x = e^a
• lim(x→0) (e^x − 1 − x)/x² = 1/2

Trigonometric limits:

• lim(x→0) (sin ax)/bx = a/b
• lim(x→0) (1 − cos x)/x² = 1/2
• lim(x→0) (tan x − x)/x³ = 1/3

Types of indeterminate forms and methods:

• 0/0: factor, rationalise, or L’Hôpital
• ∞/∞: divide by highest power or use L’Hôpital
• 0·∞: convert to 0/0 or ∞/∞ form (move denominator up or down)
• ∞ − ∞: combine into single fraction

Sandwich theorem: If f(x) ≤ g(x) ≤ h(x) and lim f = lim h = L, then lim g = L Used for trigonometric limits where direct substitution is difficult

• Key formula: lim(x→0) sin x/x = 1; lim(x→0) (e^x − 1)/x = 1; lim(x→0) (1+x)^(1/x) = e
• Common trap: sin x/x limit is only valid when x is in radians — degrees give different result!
• Exam weight: 1 question per year (4 marks); forms foundation for differentiation

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🔴 Extended — Deep Dive

Comprehensive coverage for students on a longer study timeline.

Limits — Comprehensive JEE Main Notes

Evaluating lim(x→0) sin(sin x)/x: Use series: sin x ≈ x − x³/6 + …; sin(sin x) ≈ sin(x − x³/6) ≈ (x − x³/6) − (x − x³/6)³/6 ≈ x − x³/2 + … So sin(sin x)/x ≈ 1 − x²/2 → limit = 1

General exponential limit: lim(x→a) (f(x))^{g(x)} = e^{lim(x→a) g(x)·ln(f(x))} For 1^∞ type: lim(1 + u)^{1/u} = e when u → 0

Nth root trick: lim(n→∞) n[(1 + x/n)^n − 1] = nx (use expansion) For n → ∞, (1 + x/n)^n → e^x

Using series expansions: e^x = 1 + x + x²/2! + x³/3! + … ln(1+x) = x − x²/2 + x³/3 − … sin x = x − x³/3! + x⁵/5! − … cos x = 1 − x²/2! + x⁴/4! − … tan x = x + x³/3 + 2x⁵/15 + …

Evaluating forms like ∞^0 and 0^0: Take natural log: let L = lim f^g = e^{lim g ln f} Then evaluate lim g ln f — often becomes easier form Example: lim(x→0+) x^x = e^{lim x ln x} = e^{lim ln x/(1/x)} = e^{lim (−x²/x)} = e^0 = 1

Stolz-Cesàro theorem (for sequences): If b_n is strictly monotonic and unbounded, then: lim a_n/b_n = lim (a_{n+1} − a_n)/(b_{n+1} − b_n) if the right limit exists

Using continuity: If f is continuous at a, then lim(x→a) f(x) = f(a) Example: lim(x→0) e^{sin x} = e^{lim(x→0) sin x} = e⁰ = 1

Parametric limits: lim(x→0) (e^{ax} − 1)/sin(bx) = a/b (using standard forms) lim(x→0) (ln(1 + ax))/x = a

Important limits to remember:

• lim(x→0) (sin x − x)/x³ = −1/6
• lim(x→0) (cos x − 1 + x²/2)/x⁴ = 1/24
• lim(x→0) (tan x − sin x)/x³ = 1/2
• lim(x→0) (e^x − 1 − x − x²/2)/x³ = 1/6

Limit of (1 + ax)^b/x: lim(x→0) [(1 + ax)^b − 1]/x = ab

Understanding one-sided limits: LHL = lim(x→a⁻) f(x); RHL = lim(x→a⁺) f(x) For limit to exist: both must exist and be equal Example: |x|/x as x → 0: LHL = −1, RHL = +1 → limit does not exist

Infinite limits: lim(x→∞) x^n/e^x = 0 for any n (exponential beats polynomial) lim(x→∞) ln x/x = 0 (logarithm grows slower than polynomial) lim(x→0+) x ln x = 0 (x ln x → 0 as x → 0+)

Evaluating using substitution: For lim(x→∞) tan(π/x)/sin(2/x): set u = 1/x → as x → ∞, u → 0 = tan(πu)/sin(2u) → (πu)/(2u) = π/2

• Remember: sin x/x → 1 as x → 0 (radians!); (e^x − 1)/x → 1; (1 + 1/x)^x → e; convert 1^∞ to e^{lim g·(f−1)}; use L’Hôpital for 0/0 and ∞/∞ only after confirming the form
• Previous years: “lim(x→0) (e^x − cos x)/sin x” [2023]; “lim(n→∞) [(n+1)(n+2)]^½ − n” [2024]; “lim(x→0) (sin x − x cos x)/(x³)” [2024]

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📊 JEE Main Exam Essentials

Detail	Value
Questions	90 (30 per subject)
Time	3 hours
Marks	300 (90 per subject)
Section	Physics (30), Chemistry (30), Mathematics (30)
Negative	−1 for wrong answer
Mode	Computer-based

🎯 High-Yield Topics for JEE Main Mathematics

• Calculus (Differentiation + Integration) — ~35 marks combined
• Coordinate Geometry (straight lines, circles, conics) — ~20 marks
• Algebra (Complex Numbers, Quadratics, P&C, Probability) — ~25 marks
• Trigonometry + Inverse Trigonometry — ~15 marks
• Vector + 3D — ~15 marks

📝 Previous Year Question Patterns

• Limits: 1 question per year, 4 marks
• Common patterns: evaluate limit of trigonometric expression, exponential limits, using L’Hôpital’s rule, sandwich theorem
• Weight: medium frequency, foundational for calculus

💡 Pro Tips

• Always check if the form is indeterminate before applying any technique
• L’Hôpital’s rule can be applied repeatedly if needed
• For sin x/x type, make sure x is in radians (JEE always uses radians for such limits)
• When you see 1^∞, rewrite as e^{lim (f−1)g} and then evaluate
• For rational functions at infinity, divide by highest power of x in denominator
• When factorising doesn’t work, try rationalising (multiply by conjugate)
• Series expansions are very powerful for limits — remember the first 3–4 terms of e^x, sin x, cos x, ln(1+x), (1+x)^n

🔗 Official Resources

• NTA Official JEE Main
• JEE Main Syllabus PDF

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