How to Solve: Limits (L’Hôpital’s Rule, Series Expansion, Sandwich Theorem) – IIT JEE Guide

How to Solve: Limits (L’Hôpital’s Rule, Series Expansion, Sandwich Theorem) – IIT JEE Guide

Introduction

Mastering limits with L’Hôpital’s Rule, series expansion, and the Sandwich Theorem can boost your IIT JEE score by 10-15 marks—because these techniques turn impossible-looking limits into easy, solvable problems. Whether it’s a 0/0, ∞/∞, or a tricky trigonometric limit, this guide will make you fast, accurate, and exam-ready.

What You Need To Know First

Before diving in, you must already understand: 1. Basic limit evaluation (direct substitution, factoring, rationalization). 2. Differentiation rules (product rule, quotient rule, chain rule). 3. Taylor/Maclaurin series expansions (for common functions like sin x, cos x, eˣ, ln(1+x)).

If any of these are shaky, stop now and review them first.

Key Vocabulary

Formulas To Know

1. L’Hôpital’s Rule

Formula: If lim(x→a) f(x)/g(x) = 0/0 or ∞/∞, then: lim(x→a) f(x)/g(x) = lim(x→a) f’(x)/g’(x) (Apply repeatedly if needed.)

Variables: - f(x), g(x) = differentiable functions. - a = point where limit is evaluated (can be ±∞).

MEMORISE THIS (not always given on exam sheet).

2. Common Series Expansions (Maclaurin Series)

MEMORISE THESE (given on JEE Advanced sheet, but know them cold):

When to use: When the limit has a 0/0 or ∞/∞ form and L’Hôpital’s Rule leads to messy derivatives.

3. Sandwich Theorem (Squeeze Theorem)

Formula: If g(x) ≤ f(x) ≤ h(x) for all x near a (except possibly at a), and: lim(x→a) g(x) = lim(x→a) h(x) = L, then lim(x→a) f(x) = L.

MEMORISE THIS (not always given).

Step-by-Step Method

Step 1: Check for Direct Substitution

• Plug in the value. If you get a number, you’re done.
• If you get 0/0, ∞/∞, 0×∞, ∞-∞, 1^∞, 0⁰, ∞⁰, it’s indeterminate → move to Step 2.

Step 2: Try Algebraic Simplification

• Factor, rationalize, or combine fractions.
• Example: lim(x→2) (x² - 4)/(x - 2) → factor to (x+2)(x-2)/(x-2) → x+2 → 4.

Step 3: Apply L’Hôpital’s Rule (if 0/0 or ∞/∞)

• Differentiate numerator and denominator separately.
• Re-evaluate the limit.
• Repeat if still indeterminate.

Step 4: Use Series Expansion (if L’Hôpital’s is messy)

• Replace functions with their series expansions (keep enough terms to cancel the indeterminate part).
• Simplify and evaluate.

Step 5: Apply Sandwich Theorem (for oscillating or bounded functions)

• Find two functions that "squeeze" your function and have the same limit.
• Example: lim(x→0) x sin(1/x) → -|x| ≤ x sin(1/x) ≤ |x| → limit = 0.

Step 6: Check for Special Cases (1^∞, 0⁰, ∞⁰)

• Take natural log, rewrite as e^(lim ln f(x)), then apply L’Hôpital’s or series.

Worked Examples

Example 1 – Basic (L’Hôpital’s Rule)

Problem: Evaluate lim(x→0) (eˣ - 1 - x)/x²

Step 1: Direct substitution → (1 - 1 - 0)/0 = 0/0 → indeterminate. Step 2: No obvious algebraic simplification → apply L’Hôpital’s. Step 3: Differentiate numerator and denominator: - Numerator: d/dx (eˣ - 1 - x) = eˣ - 1 - Denominator: d/dx (x²) = 2x - New limit: lim(x→0) (eˣ - 1)/2x → still 0/0 → apply L’Hôpital’s again. Step 4: Differentiate again: - Numerator: d/dx (eˣ - 1) = eˣ - Denominator: d/dx (2x) = 2 - New limit: lim(x→0) eˣ / 2 = 1/2.

Answer: 1/2

What we did and why: - Used L’Hôpital’s twice because the first application still gave 0/0. - Stopped when we got a determinate form (1/2).

Example 2 – Medium (Series Expansion)

Problem: Evaluate lim(x→0) (sin x - x + x³/6)/x⁵

Step 1: Direct substitution → (0 - 0 + 0)/0 = 0/0 → indeterminate. Step 2: No algebraic simplification → try series expansion. Step 3: Replace sin x with its series: sin x = x - x³/3! + x⁵/5! - x⁷/7! + … - Numerator: (x - x³/6 + x⁵/120 - …) - x + x³/6 = x⁵/120 - x⁷/5040 + … - Denominator: x⁵ - Simplified: (x⁵/120 - x⁷/5040 + …)/x⁵ = 1/120 - x²/5040 + … Step 4: Take limit as x→0 → 1/120.

Answer: 1/120

What we did and why: - Used series expansion because L’Hôpital’s would require 5 derivatives (messy!). - Kept terms up to x⁵ to match the denominator.

Example 3 – Exam-Style (Sandwich Theorem + Trig)

Problem: Evaluate lim(x→∞) (x + sin x)/x

Step 1: Direct substitution → ∞/∞ → indeterminate. Step 2: No algebraic simplification → try Sandwich Theorem. Step 3: Note that -1 ≤ sin x ≤ 1 → -1 ≤ sin x/x ≤ 1 for x > 0. - Rewrite numerator: (x + sin x)/x = 1 + sin x/x. - Since -1/x ≤ sin x/x ≤ 1/x, and lim(x→∞) 1/x = 0, the "sin x/x" term vanishes. Step 4: By Sandwich Theorem, lim(x→∞) sin x/x = 0. - Thus, lim(x→∞) (1 + sin x/x) = 1 + 0 = 1.

Answer: 1

What we did and why: - Recognized that sin x oscillates but is bounded, so Sandwich Theorem applies. - Avoided L’Hôpital’s (which would give cos x/1 → oscillates, no limit).

Common Mistakes

Exam Traps

1-Minute Recap

"Listen up—this is your last-minute limit survival guide. First, always plug in the value. If you get 0/0 or ∞/∞, try L’Hôpital’s: differentiate top and bottom, then plug in again. If that’s messy, use series expansions—replace sin x, eˣ, etc., with their first few terms. For oscillating functions like sin(1/x), squeeze them between two functions that go to the same limit. And watch out for 1^∞—take the natural log first! Memorize the key series, and you’ll crack 90% of limit problems in under 2 minutes. Now go practice—your 15 marks are waiting!