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CBSE · Class 11 · Mathematics

Limits (Ch 12: 12.2 and 12.3)

3 concepts · 24 practice questions with answers · Free to read

Idea of a limit

From daily life

A car's speedometer shows the speed at one instant, even though speed needs time to be measured. Limits give a way: look at shorter and shorter time gaps.

Simple definition

The limit of f(x) as x approaches a, written limx→a f(x) = L, means f(x) gets as close to L as we like when x is close to a (but not necessarily equal to a). The left-hand limit uses x < a and the right-hand limit uses x > a. The limit exists only when both are equal.

Formula: lim x→a f(x) = L

Rule 1: Find by table of values — Try x on both sides of a.

  • Easy: f(x) = 2x + 1 at x → 3: values 6.8, 6.98 from the left and 7.2, 7.02 from the right; limit 7.
  • Medium: f(x) = x² at x → 2: values near 4 on both sides; limit 4.
  • Exam-type: f(x) = |x|/x at x → 0: left limit −1, right limit 1, so the limit does not exist.

Rule 2: Value versus limit — The limit does not depend on the value at a.

  • Easy: f(x) = (x² − 1)/(x − 1): f(1) is undefined, but the limit as x → 1 is 2.
  • Medium: f(x) = x + 1 for x ≠ 2 and f(2) = 10: limit as x → 2 is 3, not 10.
  • Exam-type: A function may have a limit even where it is not defined.
Remember

Limit asks "what value is f heading towards?", not "what is f at a?".

Common mistake

Taking the limit to be f(a) even when f has a jump or hole at a.

Memory clue

Limit = where it is heading, not where it is.

Practice questions

Easy
  1. 1Find lim(x→2) (3x + 1).
  2. 2Find lim(x→0) (x² + 5).
  3. 3Find lim(x→3) (x² − 2x).
Medium
  1. 4Does lim(x→0) |x|/x exist?
  2. 5f(x) = x + 2 for x ≠ 1 and f(1) = 9. Find lim(x→1) f(x).
  3. 6Find lim(x→1) (x² − 1)/(x − 1).
Challenging
  1. 7If the left-hand limit is 4 and the right-hand limit is 6, does the limit exist?
  2. 8f(x) = x for x < 2 and f(x) = 5 − x for x ≥ 2. Does lim(x→2) f(x) exist?
Show answers (try them yourself first)
  1. 17
  2. 25
  3. 33
  4. 4No: left limit −1, right limit 1
  5. 53
  6. 62
  7. 7No
  8. 8Left limit 2, right limit 3, so no

Algebra of limits

From daily life

To find the limit of a sum, compute each limit and add. The same holds for other operations.

Simple definition

If lim f = l and lim g = m (as x → a): lim (f ± g) = l ± m; lim (f·g) = l·m; lim (f/g) = l/m if m ≠ 0; lim (c·f) = c·l.

Formula: lim (f + g) = lim f + lim g

Rule 1: Use the rules — Split the expression, find each limit, combine.

  • Easy: lim(x→2) (x² + 3x) = 4 + 6 = 10.
  • Medium: lim(x→3) (x + 1)(x − 1) = 4 × 2 = 8.
  • Exam-type: lim(x→1) (2x + 1)/(x + 3) = 3/4.

Rule 2: Quotient with zero denominator — If the denominator limit is 0, simplify first.

  • Easy: lim(x→2) (x² − 4)/(x − 2) = lim (x + 2) = 4.
  • Medium: lim(x→3) (x² − 9)/(x − 3) = 6.
  • Exam-type: lim(x→1) (x² − 3x + 2)/(x − 1) = lim (x − 2) = −1.
Remember

Direct substitution works when the denominator is not 0.

Common mistake

Substituting into 0/0 and writing 0 or ∞.

Memory clue

0/0 is a puzzle: factor and cancel.

Practice questions

Easy
  1. 1Find lim(x→1) (3x² − 2x + 4).
  2. 2Find lim(x→2) (x + 5)(x − 1).
  3. 3Find lim(x→0) (2x + 3)/(x + 1).
Medium
  1. 4Find lim(x→5) (x² − 25)/(x − 5).
  2. 5Find lim(x→4) (x² − 16)/(x − 4).
  3. 6Find lim(x→2) (x² − 5x + 6)/(x − 2).
Challenging
  1. 7Find lim(x→−1) (x² + 3x + 2)/(x + 1).
  2. 8Find lim(x→3) (x³ − 27)/(x − 3).
Show answers (try them yourself first)
  1. 15
  2. 27
  3. 33
  4. 410
  5. 58
  6. 6lim (x − 3) = −1
  7. 7lim (x + 2) = 1
  8. 8lim (x² + 3x + 9) = 27

Standard limit of (xⁿ − aⁿ)/(x − a)

From daily life

Many hard-looking limits reduce to one pattern: the difference of two powers divided by the difference of the bases.

Simple definition

For any rational n: limx→a (xⁿ − aⁿ)/(x − a) = n aⁿ⁻¹. Use it to evaluate limits quickly.

Formula: lim (xⁿ − aⁿ)/(x − a) = n aⁿ⁻¹

Rule 1: Direct use — Match with the pattern.

  • Easy: lim(x→2) (x⁴ − 16)/(x − 2) = 4 · 2³ = 32.
  • Medium: lim(x→3) (x⁵ − 243)/(x − 3) = 5 · 3⁴ = 405.
  • Exam-type: lim(x→1) (x⁷ − 1)/(x − 1) = 7.

Rule 2: After a small change — Rewrite to match the pattern.

  • Easy: lim(x→0) ((1 + x)⁵ − 1)/x = 5 (put y = 1 + x, a = 1).
  • Medium: lim(x→4) (√x − 2)/(x − 4) = 1/(2√4) = 1/4.
  • Exam-type: lim(x→8) (x^(1/3) − 2)/(x − 8) = (1/3) · 8^(−2/3) = 1/12.
Remember

The exponent comes down, the power drops by one.

Common mistake

Using the formula when the powers differ in the numerator.

Memory clue

Power comes down, power minus one.

Practice questions

Easy
  1. 1Find lim(x→2) (x³ − 8)/(x − 2).
  2. 2Find lim(x→1) (x¹⁰ − 1)/(x − 1).
  3. 3Find lim(x→3) (x⁴ − 81)/(x − 3).
Medium
  1. 4Find lim(x→0) ((1 + x)⁶ − 1)/x.
  2. 5Find lim(x→9) (√x − 3)/(x − 9).
  3. 6Find lim(x→2) (x⁷ − 128)/(x − 2).
Challenging
  1. 7Find lim(x→1) (√x − 1)/(x − 1).
  2. 8Find lim(x→2) (x⁵ − 32)/(x − 2).
Show answers (try them yourself first)
  1. 13 · 4 = 12
  2. 210
  3. 34 · 27 = 108
  4. 46
  5. 51/6
  6. 67 · 64 = 448
  7. 71/2
  8. 85 · 16 = 80