1. Chapter at a Glance
A car's speedometer shows its speed at this instant — but speed is distance ÷ time, and in an instant no time passes! Calculus solves this puzzle using limits: we look at the average speed over smaller and smaller time intervals and see what value it approaches. That limiting value is the derivative. Calculus was developed by Newton and Leibniz in the 17th century; centuries earlier, mathematicians of the Kerala school (Mādhava and his successors, 14th–16th centuries) had already worked with infinite series closely linked to these ideas. This chapter introduces limits intuitively, the key standard limits, and derivatives from first principles.
- Intuitive idea of a limit; left-hand and right-hand limits.
- Algebra of limits; limits of polynomials and rational functions.
- Standard limits: (xⁿ − aⁿ)/(x − a), sin x/x, (1 − cos x)/x.
- Derivative at a point; first principle (definition).
- Algebra of derivatives: sum, difference, product, quotient rules.
- Derivatives of xⁿ, polynomials and trigonometric functions.
2. The Idea of a Limit
We write lim (x→a) f(x) = L if f(x) gets as close as we like to L when x is close enough to a (but x ≠ a). The value f(a) itself does not matter — f need not even be defined at a.
| x | 0.9 | 0.99 | 0.999 | 1 | 1.001 | 1.01 | 1.1 |
|---|---|---|---|---|---|---|---|
| (x² − 1)/(x − 1) | 1.9 | 1.99 | 1.999 | not defined | 2.001 | 2.01 | 2.1 |
So lim (x→1) (x² − 1)/(x − 1) = 2, although the function is undefined at x = 1.
Left-hand limit lim (x→a⁻) f(x) and right-hand limit lim (x→a⁺) f(x). The limit exists only if both exist and are equal.
3. Algebra of Limits
If lim f(x) and lim g(x) exist as x → a:
- lim [f(x) ± g(x)] = lim f(x) ± lim g(x)
- lim [f(x)g(x)] = lim f(x) × lim g(x)
- lim [f(x)/g(x)] = lim f(x)/lim g(x), if lim g(x) ≠ 0
- lim [k f(x)] = k lim f(x)
Polynomials: lim (x→a) p(x) = p(a). Rational functions: lim p(x)/q(x) = p(a)/q(a) if q(a) ≠ 0; if both are 0, factor out (x − a) and cancel.
Evaluate lim (x→2) (x³ − 2x² + x − 2)/(x² − 4). (Both numerator and denominator are 0 at x = 2.)
Solution: Numerator = x²(x − 2) + (x − 2) = (x − 2)(x² + 1). Denominator = (x − 2)(x + 2). Limit = (4 + 1)/4 = 5/4.
4. Standard Limits
- lim (x→a) (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹
- lim (x→0) (sin x)/x = 1 (x in radians)
- lim (x→0) (1 − cos x)/x = 0
Why sin x/x → 1: for small x > 0, comparing the areas of a triangle, a sector and a larger triangle in the unit circle gives cos x < sin x/x < 1. As x → 0, cos x → 1, so sin x/x is squeezed to 1 (the sandwich theorem).
Evaluate (a) lim (x→1) (x¹⁵ − 1)/(x¹⁰ − 1) (b) lim (x→0) sin 4x / sin 2x.
Solution: (a) Divide top and bottom by (x − 1): 15/10 = 3/2. (b) = [sin 4x/4x × 4x]/[sin 2x/2x × 2x] → 4/2 = 2.
Evaluate lim (x→0) (1 − cos 2x)/x².
Solution: 1 − cos 2x = 2sin²x → 2(sin x/x)² → 2.
5. Derivatives
The derivative of f at x is
provided the limit exists. Geometrically, f′(a) is the slope of the tangent to y = f(x) at x = a; physically, it is the instantaneous rate of change.
Find the derivative of f(x) = x² from first principles.
Solution: [(x + h)² − x²]/h = (2xh + h²)/h = 2x + h → 2x as h → 0.
Find d/dx (sin x) from first principles.
Solution: [sin(x + h) − sin x]/h = [2 cos(x + h/2) sin(h/2)]/h = cos(x + h/2) × [sin(h/2)/(h/2)] → cos x.
6. Rules and Standard Derivatives
| Rule | Formula |
|---|---|
| Sum/difference | (u ± v)′ = u′ ± v′ |
| Constant multiple | (ku)′ = ku′ |
| Product (Leibniz rule) | (uv)′ = u′v + uv′ |
| Quotient | (u/v)′ = (u′v − uv′)/v² |
| f(x) | f′(x) |
|---|---|
| c (constant) | 0 |
| xⁿ | nxⁿ⁻¹ |
| sin x | cos x |
| cos x | −sin x |
| tan x | sec²x |
| cot x | −cosec²x |
| sec x | sec x tan x |
| cosec x | −cosec x cot x |
Differentiate (a) x² sin x (b) (x + 1)/(x − 1).
Solution: (a) 2x sin x + x² cos x. (b) [(1)(x − 1) − (x + 1)(1)]/(x − 1)² = −2/(x − 1)².
7. Solved Problems (Board Pattern)
Let f(x) = x + 2 for x ≤ 1 and f(x) = 3x for x > 1. Find lim (x→1) f(x), if it exists.
Solution: LHL = 1 + 2 = 3; RHL = 3(1) = 3. Equal → limit = 3.
f(x) = kx² for x ≤ 2 and 3 for x > 2. Find k so that lim (x→2) f(x) exists.
Solution: 4k = 3 → k = 3/4.
Find the derivative of f(x) = 3x² − 5x + 7 at x = 2 and interpret it.
Solution: f′(x) = 6x − 5 → f′(2) = 7: the slope of the tangent at (2, 9) is 7.
8. Master Formula Table
| Result | Formula |
|---|---|
| Limit exists | LHL = RHL |
| Power limit | lim (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹ |
| Trig limit | lim (x→0) sin x/x = 1 |
| Cosine limit | lim (x→0) (1 − cos x)/x = 0 |
| Derivative | f′(x) = lim (h→0) [f(x + h) − f(x)]/h |
| Power rule | d/dx xⁿ = nxⁿ⁻¹ |
| Product rule | (uv)′ = u′v + uv′ |
| Quotient rule | (u/v)′ = (u′v − uv′)/v² |
9. Master Summary Table
| Type of limit | Method | Example |
|---|---|---|
| Polynomial | Substitute | lim (x→2) x² + 1 = 5 |
| 0/0 rational | Factor and cancel | (x² − 4)/(x − 2) → 4 |
| Powers | naⁿ⁻¹ | (x⁵ − 32)/(x − 2) → 80 |
| Trigonometric | sin x/x → 1 | sin 3x/x → 3 |
| Piecewise | Compare LHL, RHL | Find k |
10. Mind Map
- LHL = RHL
- Algebra of limits
- naⁿ⁻¹
- sin x/x → 1
- First principle
- Tangent slope
- Sum, product
- Quotient; trig
11. Tips and Tricks
- Always try direct substitution first; only if you get 0/0 do you need to simplify.
- For sin(kx)/x, multiply and divide by k: answer k.
- Rewrite roots and reciprocals as powers (√x = x^(1/2), 1/x² = x⁻²) before using the power rule.
- Using degrees in sin x/x → 1 (it needs radians).
- (uv)′ = u′v′ — wrong; use the product rule.
- Thinking the limit equals f(a) always — only for nice (continuous) functions.
12. Important One-liners
- A limit describes what f(x) approaches, not what f(a) is.
- The limit exists only if the left- and right-hand limits are equal.
- lim (x→0) sin x/x = 1.
- The derivative is the slope of the tangent.
- d/dx (sin x) = cos x; d/dx (cos x) = −sin x.
- The derivative of a constant is 0.
13. Exam-pattern Questions with Detailed Answers
lim (x→3) (x + 3) equals:
- 3
- 6
- 0
- 9
Show answer
Substitute: 6.
Answer: (B)
Find the derivative of x⁵ − 3x² + 7.
Show answer
5x⁴ − 6x.
Answer: 5x⁴ − 6x
Evaluate lim (x→0) sin 5x / x.
Show answer
5 × (sin 5x)/(5x) → 5.
Answer: 5
Evaluate lim (x→2) (x⁵ − 32)/(x − 2).
Show answer
5 × 2⁴ = 80.
Answer: 80
Find the derivative of 1/x from first principles.
Show answer
[1/(x + h) − 1/x]/h = −h/[hx(x + h)] = −1/[x(x + h)] → −1/x².
Answer: −1/x²
Evaluate lim (x→0) tan x / x.
Show answer
(sin x/x) × (1/cos x) → 1 × 1 = 1.
Answer: 1
Differentiate (x + sec x)(x − tan x).
Show answer
(1 + sec x tan x)(x − tan x) + (x + sec x)(1 − sec²x).
Answer: As above
Differentiate sin x / (1 + cos x).
Show answer
[cos x(1 + cos x) + sin²x]/(1 + cos x)² = (1 + cos x)/(1 + cos x)² = 1/(1 + cos x).
Answer: 1/(1 + cos x)
Evaluate lim (x→0) [√(1 + x) − 1]/x.
Show answer
Multiply by conjugate: x/[x(√(1 + x) + 1)] → 1/2.
Answer: 1/2
Evaluate lim (x→0) (cos 2x − 1)/(cos x − 1).
Show answer
cos 2x − 1 = −2sin²x; cos x − 1 = −2sin²(x/2). Ratio = sin²x/sin²(x/2) = [(sin x/x)² × x²]/[(sin(x/2)/(x/2))² × x²/4] → 4.
Answer: 4
Find the derivative of cos x from first principles.
Show answer
[cos(x + h) − cos x]/h = −2 sin(x + h/2) sin(h/2)/h = −sin(x + h/2) × [sin(h/2)/(h/2)] → −sin x.
Answer: −sin x
If f(x) = |x|/x for x ≠ 0 and f(0) = 0, show that lim (x→0) f(x) does not exist.
Show answer
For x > 0, f(x) = 1 → RHL = 1. For x < 0, f(x) = −1 → LHL = −1. LHL ≠ RHL, so the limit does not exist.
Answer: Does not exist
14. Quick Revision Summary
- Limit = value approached; exists iff LHL = RHL.
- Substitute; if 0/0, factor/rationalise; use standard limits.
- (xⁿ − aⁿ)/(x − a) → naⁿ⁻¹; sin x/x → 1; (1 − cos x)/x → 0.
- f′(x) = lim [f(x + h) − f(x)]/h = slope of tangent.
- Rules: power, sum, product, quotient; derivatives of the six trig functions.