Limits and Derivatives ISC Class 11: Limits, First Principles and Rules

What Limits and Derivatives is about

Weightage: Limits and Derivatives belongs to the unit Calculus, which CISCE weights at 8 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Intuitive idea of a limit; left- and right-hand limits

lim x→a f(x) = L means that f(x) can be made as close to L as we wish by taking x close enough to a, x ≠ a. The value f(a) itself does not matter (it may not even exist). The left-hand limit lim x→a⁻ f(x) uses x < a; the right-hand limit lim x→a⁺ f(x) uses x > a. The limit exists if and only if both one-sided limits exist and are equal. For piecewise functions, compute each side from its own formula.

Worked example

Find lim x→0 |x|/x if it exists, and find lim x→2 f(x) where f(x) = x + 1 for x < 2 and f(x) = 3x − 3 for x ≥ 2.

Try it first, then show the solutionSolution

1. |x|/x = −1 for x < 0 and 1 for x > 0: LHL = −1, RHL = 1, unequal, so no limit
2. For f: LHL = 2 + 1 = 3; RHL = 6 − 3 = 3; equal

Answer: Does not exist; 3

Algebra of limits and indeterminate forms

If lim f = l and lim g = m (both finite), then lim (f ± g) = l ± m, lim (fg) = lm, lim (kf) = kl, and lim (f/g) = l/m if m ≠ 0 (these fundamental theorems are stated, not proved, in this course). So for a polynomial, lim x→a p(x) = p(a); for a rational function, substitute if the denominator is not zero. If substitution gives 0/0 (an indeterminate form), the expression must be simplified first: factorise and cancel the common factor (x − a), or rationalise a surd by multiplying by its conjugate.

Worked example

Evaluate (i) lim x→3 (x² − 9)/(x − 3) (ii) lim x→0 (√(1 + x) − 1)/x.

Try it first, then show the solutionSolution

1. (i) 0/0: (x − 3)(x + 3)/(x − 3) = x + 3 → 6
2. (ii) 0/0: multiply by (√(1 + x) + 1): x/(x(√(1 + x) + 1)) = 1/(√(1 + x) + 1) → 1/2

Answer: 6; 1/2

The standard limit (xⁿ − aⁿ)/(x − a)

For any rational n and a > 0, lim x→a (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹. For positive integers n this follows from xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + xⁿ⁻²a + … + aⁿ⁻¹), which has n terms each tending to aⁿ⁻¹. It handles many 0/0 forms quickly; if the denominator is a different power, divide top and bottom by (x − a).

Worked example

Evaluate lim x→2 (x⁵ − 32)/(x − 2) and lim x→2 (x⁵ − 32)/(x³ − 8).

Try it first, then show the solutionSolution

1. First: 5 × 2⁴ = 80
2. Second: divide top and bottom by (x − 2): (5 × 2⁴)/(3 × 2²) = 80/12

Answer: 80; 20/3

Trigonometric limits

With x in radians, comparing the areas of a triangle, a sector and a larger triangle in a unit circle gives sin x < x < tan x for 0 < x < π/2, so cos x < (sin x)/x < 1; squeezing gives lim x→0 (sin x)/x = 1. Consequences: lim (tan x)/x = 1; lim (1 − cos x)/x = 0; lim (1 − cos x)/x² = 1/2; and lim (sin ax)/(bx) = a/b. For 1 − cos kx, use 1 − cos kx = 2 sin²(kx/2).

Worked example

Evaluate (i) lim x→0 (sin 5x)/(sin 3x) (ii) lim x→0 (1 − cos 4x)/x².

Try it first, then show the solutionSolution

1. (i) [(sin 5x)/(5x)] × [(3x)/(sin 3x)] × (5/3) → 5/3
2. (ii) 1 − cos 4x = 2 sin²2x, so 2(sin 2x/(2x))² × 4 → 8

Answer: 5/3; 8

Exponential and logarithmic limits

Using the series for eˣ (or the definition of e), lim x→0 (eˣ − 1)/x = 1 and lim x→0 log(1 + x)/x = 1 (natural logarithm). More generally, lim x→0 (aˣ − 1)/x = log a for a > 0. Scale the variable to match: (e^{kx} − 1)/x = k × (e^{kx} − 1)/(kx) → k.

Worked example

Evaluate (i) lim x→0 (e^{3x} − 1)/x (ii) lim x→0 (2ˣ − 1)/x (iii) lim x→0 log(1 + 4x)/x.

Try it first, then show the solutionSolution

1. (i) 3 × (e^{3x} − 1)/(3x) → 3
2. (ii) log 2 ≈ 0.693
3. (iii) 4 × log(1 + 4x)/(4x) → 4

Answer: 3; log 2; 4

Derivative as rate of change and slope of tangent

For a moving object with distance s(t), the average velocity over [t, t + h] is [s(t + h) − s(t)]/h; letting h → 0 gives the instantaneous velocity s′(t). Geometrically, [f(a + h) − f(a)]/h is the slope of the secant through (a, f(a)) and (a + h, f(a + h)); as h → 0 the secant turns into the tangent, so f′(a) = lim h→0 [f(a + h) − f(a)]/h is the slope of the tangent at x = a. The tangent at (a, f(a)) is y − f(a) = f′(a)(x − a).

Worked example

A body moves s = 5t² metres in t seconds. Find its velocity at t = 3 s from the definition.

Try it first, then show the solutionSolution

1. [s(3 + h) − s(3)]/h = [5(9 + 6h + h²) − 45]/h = 30 + 5h
2. As h → 0, this tends to 30

Answer: 30 m s⁻¹

Differentiation from first principles

First principles means using f′(x) = lim h→0 [f(x + h) − f(x)]/h directly. Typical steps: expand or combine f(x + h) − f(x); factor out h (using algebra, rationalisation or a sum-to-product formula); cancel h; let h → 0. Results: d/dx(xⁿ) = nxⁿ⁻¹; d/dx(√x) = 1/(2√x); d/dx(1/x) = −1/x²; d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(constant) = 0.

Worked example

Find the derivative of √x from first principles.

Try it first, then show the solutionSolution

1. [√(x + h) − √x]/h × [√(x + h) + √x]/[√(x + h) + √x] = h/(h[√(x + h) + √x])
2. = 1/(√(x + h) + √x) → 1/(2√x) as h → 0

Answer: 1/(2√x)

Sum, product and quotient rules

(u ± v)′ = u′ ± v′; (ku)′ = ku′; (uv)′ = u′v + uv′ (product rule); (u/v)′ = (u′v − uv′)/v², v ≠ 0 (quotient rule: 'low d-high minus high d-low, over low squared'). With these and the basic derivatives you get d/dx(tan x) = sec²x, d/dx(cot x) = −cosec²x, d/dx(sec x) = sec x tan x and d/dx(cosec x) = −cosec x cot x. The derivative of a product is not the product of the derivatives.

Worked example

Differentiate (i) x² sin x (ii) (x + 1)/(x − 1).

Try it first, then show the solutionSolution

1. (i) Product rule: 2x sin x + x² cos x
2. (ii) Quotient rule: [1 × (x − 1) − (x + 1) × 1]/(x − 1)² = −2/(x − 1)²

Answer: 2x sin x + x² cos x; −2/(x − 1)²

Maths in real life: Limits and Derivatives

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Differentiation from First Principles and Rules of Differentiation
• Limits and Standard Limits

Limits and Derivatives formula sheet

Existence of a limit

${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=L$

Both one-sided limits must agree.

Power limit

${\lim }_{x\to a}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$

Useful for 0/0 forms with powers.

Trigonometric limits

${\lim }_{x\to 0}\frac{\sin x}{x}=1;{\lim }_{x\to 0}\frac{\tan x}{x}=1;{\lim }_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$

x in radians.

Exponential and log limits

${\lim }_{x\to 0}\frac{e^x-1}{x}=1;{\lim }_{x\to 0}\frac{a^x-1}{x}=\log a;{\lim }_{x\to 0}\frac{\log (1+x)}{x}=1$

Natural logarithm.

Definition of the derivative

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Slope of the tangent; instantaneous rate.

Basic derivatives

$(x^n{)}^{\prime }=n{x}^{n-1};(\sin x{)}^{\prime }=\cos x;(\cos x{)}^{\prime }=-\sin x;(\tan x{)}^{\prime }={\sec }^{2}x$

(sec x)′ = sec x tan x.

Product rule

$(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$

Not u′v′.

Quotient rule

${\left(\frac{u}{v}\right)}^{\prime }=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$

Order matters in the numerator.

Limits and Derivatives: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Limits and Derivatives

The question patterns to expect, and the method that works for each.

Tricks to remember Limits and Derivatives

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Left equals right

  No agreement between LHL and RHL, no limit.

  Use it when: Existence of limits.

• 0/0 means do more work

  0/0 is not an answer; it is a signal to factorise, rationalise or use a standard limit.

  Use it when: Indeterminate forms.

• Match the angle

  sin(kx) needs kx underneath: sin 5x/(5x) → 1.

  Use it when: Trigonometric limits.

• Low d-high minus high d-low

  (u/v)′ = (v u′ − u v′)/v².

  Use it when: Quotient rule.

• Power comes down, power goes down

  d/dx(xⁿ) = nxⁿ⁻¹.

  Use it when: Power rule.

Limits and Derivatives: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Limits and Derivatives

The slips examiners see most often, and what to write instead.

• ✗Mistake: Writing 0/0 as the answer

  Students stop after substituting.

  Wrong working

  lim x→3 (x² − 9)/(x − 3) = 0/0, so the limit does not exist.

  Right working

  Cancel (x − 3): limit = 6.

  ✓Do this: 0/0 means simplify, not stop.

  Usually costs: Full marks for the part.

• ✗Mistake: Using degrees in sin x/x

  The result sin x/x → 1 holds only for radians.

  Wrong working

  lim (sin x°)/x = 1.

  Right working

  x° = πx/180 radians, so the limit is π/180.

  ✓Do this: Calculus angles are in radians unless stated.

  Usually costs: 1 mark.

• ✗Mistake: Product rule as product of derivatives

  Students assume (uv)′ = u′v′.

  Wrong working

  d/dx(x² sin x) = 2x cos x.

  Right working

  2x sin x + x² cos x.

  ✓Do this: Two terms: one derivative in each.

  Usually costs: 1-2 marks.

• ✗Mistake: Reversing the quotient-rule numerator

  Order matters because of the minus sign.

  Wrong working

  d/dx[(x + 1)/(x − 1)] = [(x + 1) − (x − 1)]/(x − 1)² = 2/(x − 1)².

  Right working

  [(x − 1) − (x + 1)]/(x − 1)² = −2/(x − 1)².

  ✓Do this: 'Low d-high' first.

  Usually costs: 1 mark.

• ✗Mistake: Not letting h → 0 at the right time

  Students put h = 0 before cancelling, getting 0/0.

  Wrong working

  [(x + h)² − x²]/h at h = 0 is 0/0.

  Right working

  Simplify to 2x + h first, then let h → 0: 2x.

  ✓Do this: Cancel h, then take the limit.

  Usually costs: 1-2 marks.

Practice questions on Limits and Derivatives

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   lim x→0 (1 − cos x)/x equals:
   (a) 1
   (b) 0
   (c) 1/2
   (d) does not exist

   Hint

   Write 1 − cos x = 2 sin²(x/2).

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   The derivative of x⁵ is:
   (a) x⁴
   (b) 5x⁴
   (c) 5x⁶
   (d) x⁶/6

   Hint

   nxⁿ⁻¹.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   Evaluate lim x→3 (x² + 2x).

   Hint

   Substitute.

   Answer

   15

4. Q4 · 1 mark · Very short answerExam level

   Differentiate sin x + cos x.

   Hint

   Sum rule.

   Answer

   cos x − sin x

5. Q5 · 2 marks · Short answerExam level

   Evaluate lim x→0 (sin 4x)/(tan 2x).

   Hint

   Divide top and bottom by x.

   Answer

   4/2 = 2

6. Q6 · 2 marks · Short answerHarder exam level

   Evaluate lim x→−2 (x³ + 8)/(x + 2).

   Hint

   x³ − (−2)³ over x − (−2).

   Answer

   3(−2)² = 12

7. Q7 · 2 marks · Short answerHarder exam level

   Differentiate √x + 1/√x.

   Hint

   Write as x^{1/2} + x^{−1/2}.

   Answer

   1/(2√x) − 1/(2x√x)

8. Q8 · 3 marks · Short answerHarder exam level

   Evaluate lim x→0 (e^{2x} − 1)/(sin 3x).

   Hint

   Divide top and bottom by x.

   Answer

   [2(e^{2x} − 1)/(2x)]/[3 sin 3x/(3x)] → 2/3

9. Q9 · 2 marks · Short answerHarder exam level

   Evaluate lim x→0 (√(4 + x) − 2)/x.

   Hint

   Rationalise.

   Answer

   1/(√(4 + x) + 2) → 1/4

10. Q10 · 3 marks · Short answerHarder exam level

    Differentiate sin x/(1 + cos x) and simplify.

    Hint

    Use sin²x + cos²x = 1 in the numerator.

    Answer

    [cos x(1 + cos x) + sin²x]/(1 + cos x)² = (1 + cos x)/(1 + cos x)² = 1/(1 + cos x)

11. Q11 · 2 marks · Short answerHarder exam level

    Differentiate (2x + 3)/(x − 1).

    Hint

    Quotient rule.

    Answer

    [2(x − 1) − (2x + 3)]/(x − 1)² = −5/(x − 1)²

12. Q12 · 3 marks · Short answerHarder exam level

    Evaluate lim x→0 log(1 + 3x)/(e^{2x} − 1).

    Hint

    Divide top and bottom by x.

    Answer

    3/2

13. Q13 · 5 marks · Long answerHarder exam level

    (a) f(x) = x² for x ≤ 1 and f(x) = 2x − 1 for x > 1. Find lim x→1 f(x). [2]
    (b) Find the slope of the tangent to y = x³ − 3x at x = 2. [1]
    (c) Find the points on y = x³ − 3x where the tangent is parallel to the x-axis. [2]

    Hint

    Horizontal tangent: dy/dx = 0.

    Answer

    (a) LHL = 1, RHL = 1, so the limit is 1. (b) dy/dx = 3x² − 3 = 9. (c) 3x² − 3 = 0 ⇒ x = ±1: points (1, −2) and (−1, 2).

14. Q14 · 3 marks · Short answerHOTS / competency

    Evaluate lim x→0 (cos 2x − cos 6x)/x².

    Hint

    cos C − cos D = −2 sin((C + D)/2) sin((C − D)/2).

    Answer

    cos 2x − cos 6x = 2 sin 4x sin 2x, so the limit is 2 × 4 × 2 = 16

15. Q15 · 3 marks · Short answerHOTS / competency

    If y = (x + cos x)/(x − cos x), find dy/dx.

    Hint

    Expand the numerator of the quotient rule; most terms cancel.

    Answer

    [(1 − sin x)(x − cos x) − (x + cos x)(1 + sin x)]/(x − cos x)² = [x − cos x − x sin x + sin x cos x − x − x sin x − cos x − sin x cos x]/(x − cos x)² = −2(x sin x + cos x)/(x − cos x)²

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.When does lim x→a f(x) exist?Show answer

   When LHL and RHL exist and are equal.

2. 2.lim x→a (xⁿ − aⁿ)/(x − a)?Show answer

   naⁿ⁻¹.

3. 3.lim x→0 (sin x)/x?Show answer

   1 (x in radians).

4. 4.lim x→0 (eˣ − 1)/x?Show answer

   1.

5. 5.lim x→0 log(1 + x)/x?Show answer

   1.

6. 6.Definition of f′(x)?Show answer

   lim h→0 [f(x + h) − f(x)]/h.

7. 7.Derivative of cos x?Show answer

   −sin x.

8. 8.Quotient rule?Show answer

   (u′v − uv′)/v².

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you find LHL and RHL of a piecewise function?
• ☐Can you evaluate a 0/0 limit by factorising and by rationalising?
• ☐Can you prove lim (sin x)/x = 1?
• ☐Can you evaluate (1 − cos kx)/x² limits?
• ☐Can you evaluate (e^{kx} − 1)/x and log(1 + kx)/x limits?
• ☐Can you differentiate √x, 1/x and sin x from first principles?
• ☐Can you apply the product and quotient rules without sign errors?
• ☐Can you find the slope of a tangent and a velocity from a formula?

Limits and Derivatives: frequently asked questions