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State Board · Class 11 · Mathematics Part 2

Lesson 7: Limits

5 concepts · 40 practice questions with answers · Free to read

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Meaning of a limit

From daily life

As a lift nears the 6th floor it slows down, and before it arrives we say it is "going towards 6". The idea that tells which number the value of a function approaches is the limit. Building on classes 9–10 : Since classes 9–10 we have found the value of a function by putting in x; here we see what happens as x gets close to a number.

Simple definition

If f(x) gets close to l as x gets close to a (without being a), the limit is l, written lim f(x) = l as x → a. The limit exists only when the values from the left (x → a⁻) and from the right (x → a⁺) are equal.

Rule 1: Algebra of limits — lim (f ± g) = lim f ± lim g; lim (f g) = lim f · lim g; lim (f/g) = lim f / lim g (if the limit of the denominator is not 0); lim (c f) = c lim f.

36leftrightx → 3⁻ : 5.8, 5.9x → 3⁺ : 6.1, 6.2limit = 6

In the figure : f(x) = x + 3: as x → 3 the values from the left 5.8, 5.9 and from the right 6.1, 6.2 approach 6.

How is the formula obtained? : A limit is not the value of f at a but the behaviour near a. So the limit can exist even when f(a) does not.

  • Easy : lim (x + 3) = 6 as x → 3; both sides give 6.
  • Medium : f(x) = x + 1 (x < 2), 3x − 3 (x ≥ 2): left side 2 + 1 = 3, right side 3 × 2 − 3 = 3; the limit is 3.
  • Exam-type : f(x) = x² (x < 1), 2x + 1 (x ≥ 1): left 1, right 3; they differ, so the limit does not exist. ∣x∣/x: left −1, right 1; no limit at 0.
Remember

To find a limit x only has to get near a, never to touch it.

Common mistake

Taking the limit and f(a) as the same. The limit can exist even when f(a) does not (a graph with a hole).

Shortcut

Put x = a − 0.01 and a + 0.01 and compare; if they agree that is the limit. Memory clue : "Left, right, if they match it is the limit".

CET trap

For a function given in pieces find the left and right sides separately; the formula changes.

CET-style example : lim ∣x∣/x as x → 0 is? (A) 1 (B) −1 (C) 0 (D) does not exist

Answer : (D). Left −1, right 1. Time-saving method : Check both sides; different means no limit.

Spot the difference

Pointf(a)Limit
Meaningthe value at athe behaviour near a
May fail to exist?yes (a hole)yes (left ≠ right)
xis put equal to aonly approaches a

Practice questions

Easy
  1. 1Find lim (2x + 1) as x → 5.
  2. 2Find lim (x² + 1) as x → 2.
  3. 3Why can x = a not always be put in to find a limit?
Medium
  1. 4f(x) = 3x − 1 (x < 2), x + 3 (x ≥ 2): find the left side as x → 2.
  2. 5Find the right side of the same function.
  3. 6Find lim (2x + 1) as x → 0.
Challenging
  1. 7Does ∣x − 3∣/(x − 3) have a limit as x → 3?
  2. 8Find lim (x² − 25)/(x − 5) as x → 5.
Show answers (try them yourself first)
  1. 111.
  2. 25.
  3. 3f(a) अस्तित्वात नसेल (उदा. 0/0) तर.
  4. 43 × 2 − 1 = 5.
  5. 52 + 3 = 5.
  6. 61.
  7. 7नाही ; डावी −1, उजवी 1.
  8. 8(x + 5) ; 10.

Methods of evaluating limits

From daily life

A 0/0 blocks the answer like a jammed door. Factorising or multiplying by the conjugate of a root removes the block. Building on classes 9–10 : Since classes 9–10 we know factorising and rationalising surds; here we use the same tools.

Simple definition

Methods: (1) direct substitution; (2) for 0/0 factorise and cancel the common factor; (3) with a square root multiply by the conjugate; (4) the formula lim (xⁿ − aⁿ)/(x − a) = n aⁿ⁻¹.

Rule 1: Order of work — First try x = a. Only for 0/0 use the next method; for another form such as k/0 the limit is infinite or does not exist.

24(x² − 4)/(x − 2)= x + 2 (x ≠ 2)hole at (2, 4)limit = 4

In the figure : (x² − 4)/(x − 2) = x + 2: a hole at (2, 4); the limit is 4.

How is the formula obtained? : For x ≠ 2, (x² − 4)/(x − 2) = (x + 2)(x − 2)/(x − 2) = x + 2; in a limit x never reaches 2, so cancelling (x − 2) is valid.

  • Easy : lim (x² + 3x − 4)/(x − 1) as x → 1: (x + 4)(x − 1)/(x − 1) = x + 4 → 5.
  • Medium : lim (√(x + 4) − 2)/x as x → 0: multiply top and bottom by √(x + 4) + 2 → x/(x(√(x + 4) + 2)) → 1/4.
  • Exam-type : lim (x⁵ − 2⁵)/(x − 2) as x → 2 is 5 × 2⁴ = 80; lim (x³ − 8)/(x² − 4) = 3 · 2²/(2 · 2) = 3.
Remember

In a limit x is not at a, so the factor (x − a) can be cancelled.

Common mistake

Taking 0/0 as 0 or 1. It is an indeterminate form; simplify further.

Shortcut

On seeing (xⁿ − aⁿ)/(x − a) write n aⁿ⁻¹ at once; do not waste time factorising. Memory clue : "Bring the power down, one less".

CET trap

The conjugate of √(x + 4) − 2 is √(x + 4) + 2; write it exactly.

CET-style example : lim (x³ − 8)/(x − 2) as x → 2 is? (A) 12 (B) 8 (C) 6 (D) 3

Answer : (A). 3 × 2² = 12. Time-saving method : Use n aⁿ⁻¹ with n = 3, a = 2.

Spot the difference

FormMethodExample
0/0 polynomialfactorise(x² − 4)/(x − 2) → 4
0/0 with a rootmultiply by the conjugate(√(x + 4) − 2)/x → 1/4
xⁿ − aⁿformula n aⁿ⁻¹(x³ − 8)/(x − 2) → 12

Practice questions

Easy
  1. 1Find lim (2x + 1) as x → 5.
  2. 2Find lim (x² − 9)/(x − 3) as x → 3.
  3. 3Find lim (x² + 1) as x → 2.
Medium
  1. 4Find lim (x³ − 1)/(x − 1) as x → 1.
  2. 5Find lim (√(x + 9) − 3)/x as x → 0.
  3. 6Find lim (x² − 4)/(x + 2) as x → −2.
Challenging
  1. 7Find lim (√x − 2)/(x − 4) as x → 4.
  2. 8Find lim (x⁴ − 1)/(x² − 1) as x → 1.
Show answers (try them yourself first)
  1. 111.
  2. 2x + 3 → 6.
  3. 35.
  4. 43 × 1² = 3.
  5. 51/(√(x + 9) + 3) → 1/6.
  6. 6x − 2 → −4.
  7. 71/(√x + 2) → 1/4.
  8. 8x² + 1 → 2.

Limits of trigonometric functions

From daily life

For a small angle the length of the arc and the length of the sine line look almost equal. So sin θ/θ gets close to 1. Building on classes 9–10 : Since classes 9–10 we know the ratios sin, cos, tan; here we see how they behave for a very small angle.

Simple definition

If θ is in radians, lim sin θ/θ = 1 as θ → 0. From it lim tan θ/θ = 1, lim (1 − cos θ)/θ² = 1/2, lim sin kx/x = k.

Rule 1: Adjusting the form — For sin 3x/x make the denominator 3x: (sin 3x/3x) × 3 → 3. The angle and the denominator must be the same. 1 − cos θ = 2 sin²(θ/2).

θsin θtan θsin θ < θ < tan θdivide by sin θsin θ/θ → 1θ in radians

In the figure : Unit circle: sin θ < θ < tan θ; sin θ/θ → 1.

How is the formula obtained? : For small θ, sin θ < θ < tan θ. Dividing by sin θ gives 1 < θ/sin θ < 1/cos θ; as θ → 0, cos θ → 1, so θ/sin θ → 1.

  • Easy : lim sin 3x/x = 3; lim tan 5x/x = 5.
  • Medium : lim (1 − cos x)/x² = lim 2 sin²(x/2)/x² = 2 · (1/4) = 1/2.
  • Exam-type : lim sin 3x/sin 5x = (sin 3x/3x) · (5x/sin 5x) · (3/5) → 3/5.
Remember

The formula holds only in radians; convert degrees first (180° = π).

Common mistake

Writing 1 for sin 3x/x. The angle and denominator differ; the answer is 3.

Shortcut

sin kx/mx → k/m and tan kx/mx → k/m; (1 − cos kx)/x² → k²/2. Memory clue : "Angle multiplier on top, denominator multiplier below".

CET trap

If x does not tend to 0 (for example x → π) the formula does not apply directly; put t = x − π first.

CET-style example : lim sin 7x/(3x) as x → 0 is? (A) 7/3 (B) 3/7 (C) 7 (D) 1

Answer : (A). (7/3)(sin 7x/7x) → 7/3. Time-saving method : Top multiplier divided by bottom multiplier: 7/3.

Spot the difference

LimitAnswerReason
sin x/x1basic formula
tan x/x1sin x/x · 1/cos x
(1 − cos x)/x²1/22 sin²(x/2)
sin kx/xkmatch angle and denominator

Practice questions

Easy
  1. 1Find lim sin 2x/x.
  2. 2Find lim tan x/x.
  3. 3Find lim sin 4x/(2x).
Medium
  1. 4Find lim sin 3x/sin 4x.
  2. 5Find lim (1 − cos 2x)/x².
  3. 6Find lim tan 3x/x.
Challenging
  1. 7Find lim (1 − cos x)/(x sin x).
  2. 8Find lim sin²3x/x².
Show answers (try them yourself first)
  1. 12.
  2. 21.
  3. 34/2 = 2.
  4. 43/4.
  5. 52² / 2 = 2.
  6. 63.
  7. 7(x²/2)/(x · x) = 1/2.
  8. 83² = 9.

Limits of exponential and logarithmic functions

From daily life

When a bank balance grows continuously, endless tiny increases give one fixed number e. Here we see the limits tied to the rate of such growth. Building on classes 9–10 : Since classes 9–10 we know the laws of indices and logarithms; here are their limits.

Simple definition

lim (aˣ − 1)/x = log a; in particular lim (eˣ − 1)/x = 1. lim log (1 + x)/x = 1. lim (1 + x)¹ᐟˣ = e as x → 0, e ≈ 2.718.

Rule 1: Matching the form — For e²ˣ − 1 make the denominator 2x too: (e²ˣ − 1)/2x × 2 → 2. log (1 + 3x)/x → 3. (1 + 2x)¹ᐟˣ → e².

f(x) = (2ˣ − 1)/xx = 0.10.7177x = 0.010.6956x = 0.0010.6934x = 0.00010.6932limit = log 2 = 0.6931general: (aˣ − 1)/x → log a

In the figure : (2ˣ − 1)/x: as x shrinks 0.7177, 0.6956, 0.6934, 0.6932 → log 2.

How is the formula obtained? : e is defined by the limit of (1 + x)¹ᐟˣ. Taking its log gives log (1 + x)/x → 1; and writing aˣ = e^(x log a) gives (aˣ − 1)/x → log a.

  • Easy : lim (e²ˣ − 1)/x = 2; lim (2ˣ − 1)/x = log 2 ≈ 0.6931.
  • Medium : lim log (1 + 3x)/x = 3; lim (1 + x)¹ᐟˣ = e ≈ 2.718.
  • Exam-type : lim (e³ˣ − 1)/sin 2x = (e³ˣ − 1)/3x · (2x/sin 2x) · (3/2) → 3/2.
Remember

Here log means the natural logarithm with base e; log e = 1.

Common mistake

Writing 1 for (aˣ − 1)/x. It is 1 only for a = e; otherwise log a.

Shortcut

Take the multiplier k out: (e^kx − 1)/x → k; log (1 + kx)/x → k. Memory clue : "The multiplier in the power comes into the answer".

CET trap

(1 + kx)¹ᐟˣ → eᵏ, not e.

CET-style example : lim (e⁵ˣ − 1)/x as x → 0 is? (A) 5 (B) 1 (C) e⁵ (D) 0

Answer : (A). (e⁵ˣ − 1)/5x × 5 → 5. Time-saving method : The 5 in the power is the answer.

Spot the difference

Limit (x → 0)Answer
(eˣ − 1)/x1
(aˣ − 1)/xlog a
log (1 + x)/x1
(1 + x)¹ᐟˣe

Practice questions

Easy
  1. 1Find lim (eˣ − 1)/x.
  2. 2Find lim log (1 + x)/x.
  3. 3Find lim (e⁴ˣ − 1)/x.
Medium
  1. 4Find lim (3ˣ − 1)/x (four decimals).
  2. 5Find lim log (1 + 5x)/x.
  3. 6Find lim (1 + 2x)¹ᐟˣ (e² ≈ 7.389).
Challenging
  1. 7Find lim (e²ˣ − 1)/(3x).
  2. 8Find lim (2ˣ − 1)/(3ˣ − 1) (four decimals).
Show answers (try them yourself first)
  1. 11.
  2. 21.
  3. 34.
  4. 4log 3 ≈ 1.0986.
  5. 55.
  6. 6e² ≈ 7.389.
  7. 72/3.
  8. 8log 2/log 3 ≈ 0.6309.

Limits at infinity and infinite limits

From daily life

If a tank is filled for longer and longer, the water gets close to its capacity. A limit at infinity tells where the value of a function goes when x becomes very large. Building on classes 9–10 : Since classes 9–10 we have divided by large numbers; dividing 1 by a large number gives a small result.

Simple definition

lim 1/x = 0 as x → ∞. Infinite limit: for x → 0⁺, 1/x → +∞, and for x → 0⁻, 1/x → −∞. For a ratio of polynomials divide the top and bottom by the highest power.

Rule 1: Comparing degrees — If the top has the smaller degree the limit is 0; if the degrees are equal it is the ratio of the leading coefficients; if the top has the larger degree it is infinite.

y → 0x → 0⁺ : y → ∞x → ∞ : 1/x → 0lim 1/x = 0 as x → ∞

In the figure : 1/x: as x grows y → 0; as x → 0⁺ y grows without bound.

How is the formula obtained? : Divide (3x² + 2x)/(5x² − 1) by x²: (3 + 2/x)/(5 − 1/x²). As x → ∞, 2/x and 1/x² go to 0; the answer is 3/5.

  • Easy : lim (2x + 1)/(x − 3) as x → ∞ = (2 + 1/x)/(1 − 3/x) → 2.
  • Medium : lim (3x² + 2x)/(5x² − 1) → 3/5; lim (x² + 1)/(x³ + x) → 0.
  • Exam-type : lim √(x² + 1)/x as x → +∞ = √(1 + 1/x²) → 1; lim (x² + 1)/(x + 1) → ∞ (the top has the larger degree).
Remember

Infinity is not a number; "the limit is ∞" means the value grows without bound.

Common mistake

Taking ∞/∞ as 1. First divide by the highest power.

Shortcut

With equal degrees write only the ratio of leading coefficients: (4x² + 3)/(2x² − x) → 4/2. Memory clue : "The biggest power wins".

CET trap

For x → −∞, √(x²) = ∣x∣ = −x; watch the sign.

CET-style example : lim (4x² + 3)/(2x² − x) as x → ∞ is? (A) 2 (B) 4 (C) 1/2 (D) ∞

Answer : (A). 4/2. Time-saving method : Equal degrees: ratio of leading coefficients.

Spot the difference

Top : bottom (degree)Limit (x → ∞)Example
smaller : larger0(x² + 1)/(x³ + x)
equalratio of coefficients(3x² + 2x)/(5x² − 1) → 3/5
larger : smaller±∞(x² + 1)/(x + 1)

Practice questions

Easy
  1. 1Find lim 5/x as x → ∞.
  2. 2Find lim (2x + 1)/(x + 4) as x → ∞.
  3. 3Find lim 1/x² as x → ∞.
Medium
  1. 4Find lim (3x² + x)/(6x² − 5) as x → ∞.
  2. 5Find lim (x + 1)/(x² + 3) as x → ∞.
  3. 6What happens to lim 1/x as x → 0⁺?
Challenging
  1. 7Find lim (2x³ + x)/(5x³ − 4x²) as x → ∞.
  2. 8Find lim √(x² + 4)/(2x) as x → +∞.
Show answers (try them yourself first)
  1. 10.
  2. 22.
  3. 30.
  4. 43/6 = 1/2.
  5. 50.
  6. 6+∞ (किंमत मर्यादेशिवाय वाढते).
  7. 72/5.
  8. 81/2.

Lesson 7: Limits — quick revision

TopicFormula / point
Limitexists only if left = right; f(a) need not exist
0/0factorise; multiply by the conjugate; (xⁿ − aⁿ)/(x − a) → n aⁿ⁻¹
Trigonometricsin x/x → 1; tan x/x → 1; (1 − cos x)/x² → 1/2; sin kx/x → k
Exponential(aˣ − 1)/x → log a; (eˣ − 1)/x → 1; log (1 + x)/x → 1; (1 + x)¹ᐟˣ → e
At infinity1/x → 0; equal degrees: ratio of coefficients; larger top degree: infinite

🎯 Most important points for the exam

  • First try x = a.
  • The formulas hold only in radians.
  • Match the angle and the denominator.
  • The multiplier in the power comes into the answer.
  • Divide by the highest power.
Study mantra

"Left, right, if they match it is the limit".