Calculus: Limits, Differentiation & Integration
What it is
Calculus studies how quantities change. A limit describes the value a function approaches as its input nears some point, even where the function itself is undefined there. Continuity extends this: a function is continuous at a point when its limit there equals its actual value, so the graph has no break or jump. Differentiation measures the instantaneous rate of change — the slope of the curve at a single point — and from it come tangent lines, normal lines, and the tests for a graph's highest and lowest points. Integration reverses differentiation: given a rate of change, it recovers the original quantity, and a definite integral further computes the signed area swept out under a curve between two limits. Together these tools answer two everyday questions about any function: how fast is it changing right now, and how much has it accumulated over an interval.
Core concepts
Limits and continuity. The limit of f(x) as x approaches a value "a", written lim(x→a) f(x), is the value f(x) nears as x nears a. Many limits that look undefined (a 0 ÷ 0 form) simplify by factoring first: for f(x) = (x² − 4) ÷ (x − 2), direct substitution at x = 2 gives 0 ÷ 0, but factoring gives (x − 2)(x + 2) ÷ (x − 2) = x + 2 for every x ≠ 2, so the limit as x → 2 is 2 + 2 = 4, even though f(2) is undefined. A function is continuous at a point when the limit from both sides equals its actual value there — no break in the graph.
Standard derivatives.
| Function | Derivative |
|---|---|
| xⁿ | n × xⁿ⁻¹ |
| sin x | cos x |
| cos x | −sin x |
| eˣ | eˣ |
| ln x | 1 ÷ x |
Combining rules. For functions u(x) and v(x): the product rule gives d(uv)/dx = u'v + uv'; the quotient rule gives d(u ÷ v)/dx = (u'v − uv') ÷ v²; the chain rule differentiates a composite function outside-in, then multiplies by the derivative of the inside: d/dx[f(g(x))] = f'(g(x)) × g'(x).
Tangents and normals. At a point (x₀, f(x₀)) on a curve, the tangent line touches the curve there without crossing it and has slope f'(x₀), giving y − f(x₀) = f'(x₀) × (x − x₀). The normal is perpendicular to the tangent at that point, with slope −1 ÷ f'(x₀) (when f'(x₀) ≠ 0), giving y − f(x₀) = [−1 ÷ f'(x₀)] × (x − x₀).
Maxima and minima. A stationary point satisfies the first-order condition f'(x) = 0. The second derivative classifies it: f''(x) < 0 marks a local maximum (curving downward, like a hilltop); f''(x) > 0 marks a local minimum (curving upward, like the bottom of a bowl). This sign pairing is fixed — never the reverse.
Integration — building blocks.
| Integral | Result | ||
|---|---|---|---|
| ∫xⁿ dx (n ≠ −1) | xⁿ⁺¹ ÷ (n + 1) + C | ||
| ∫1 ÷ x dx | ln\ | x\ | + C |
| ∫eˣ dx | eˣ + C | ||
| ∫sin x dx | −cos x + C | ||
| ∫cos x dx | sin x + C |
Integration by parts. For a product of two functions, ∫u dv = uv − ∫v du — choose u as the factor that simplifies on differentiating, and dv as the rest.
Integration by substitution. When the integrand contains a function and, up to a constant multiple, its own derivative, set u equal to the inner function; du then replaces the matching piece, turning the integral into a simpler one in u, converted back to x once integrated.
Integration by partial fractions. A proper rational function (numerator's degree below the denominator's) with a factorable denominator can be split into simpler fractions, each easy to integrate on its own, before any integration is attempted. For distinct linear factors, each factor (x − a) contributes a term A ÷ (x − a), and the unknown constants are found by clearing denominators and matching the resulting equation.
Definite integrals and area. The Fundamental Theorem of Calculus states that if F is any antiderivative of f (that is, F' = f), then the definite integral from a to b is ∫[a to b] f(x) dx = F(b) − F(a). This quantity is the signed area between the curve and the x-axis over [a, b] — area above the axis counts positive, area below counts negative.
Worked example
- Differentiation and stationary points. Let f(x) = x³ − 6x² + 9x + 2. Then f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3), so the stationary points are x = 1 and x = 3. The second derivative is f''(x) = 6x − 12. At x = 1: f''(1) = 6 − 12 = −6 < 0, a local maximum, with f(1) = 1 − 6 + 9 + 2 = 6. At x = 3: f''(3) = 18 − 12 = 6 > 0, a local minimum, with f(3) = 27 − 54 + 27 + 2 = 2.
- Tangent and normal. Using the same f, at x₀ = 2: f(2) = 8 − 24 + 18 + 2 = 4, and f'(2) = 3(4) − 12(2) + 9 = 12 − 24 + 9 = −3. The tangent is y − 4 = −3(x − 2), i.e. y = −3x + 10. The normal has slope −1 ÷ (−3) = 1 ÷ 3, giving y − 4 = (1 ÷ 3)(x − 2), i.e. y = (1 ÷ 3)x + 10 ÷ 3.
- Definite integral and area (Fundamental Theorem). For f(x) = 2x + 3 on [0, 1]: an antiderivative is F(x) = x² + 3x, so ∫[0 to 1] (2x + 3) dx = F(1) − F(0) = (1 + 3) − 0 = 4. A trapezoid check confirms it directly: the line runs from height 3 at x = 0 to height 5 at x = 1, and a trapezoid of parallel sides 3 and 5 over width 1 has area (3 + 5) ÷ 2 × 1 = 4.
- Integration by parts. ∫x eˣ dx: take u = x, dv = eˣ dx, so du = dx, v = eˣ. Then ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C = (x − 1)eˣ + C. Check by differentiating the result: d/dx[(x − 1)eˣ] = eˣ + (x − 1)eˣ = eˣ[1 + (x − 1)] = x eˣ, which matches the original integrand.
- Integration by substitution. ∫2x(x² + 1)⁴ dx: let u = x² + 1, so du = 2x dx. The integral becomes ∫u⁴ du = u⁵ ÷ 5 + C = (x² + 1)⁵ ÷ 5 + C. Check: d/dx[(x² + 1)⁵ ÷ 5] = (1 ÷ 5) × 5(x² + 1)⁴ × 2x = 2x(x² + 1)⁴, matching the integrand.
- Integration by partial fractions. 1 ÷ [(x + 1)(x + 2)] = A ÷ (x + 1) + B ÷ (x + 2). Clearing denominators: 1 = A(x + 2) + B(x + 1). Setting x = −1 gives 1 = A(1), so A = 1; setting x = −2 gives 1 = B(−1), so B = −1. So 1 ÷ [(x + 1)(x + 2)] = 1 ÷ (x + 1) − 1 ÷ (x + 2), and ∫1 ÷ [(x + 1)(x + 2)] dx = ln|x + 1| − ln|x + 2| + C.
Common traps
- Swapping the maxima/minima sign rule — writing f''(x) > 0 for a maximum. Anchor it to shape: a maximum is the top of a hill, curving downward, so its second derivative is negative there.
- Forgetting the constant of integration, C, on an indefinite integral — a definite integral needs no C since it cancels in F(b) − F(a), but an indefinite integral without C is incomplete.
- Using the normal's slope formula backward — the normal's slope is −1 ÷ f'(x₀), the negative reciprocal, not simply −f'(x₀).
- Applying the power rule to ∫1 ÷ x dx by writing x⁻¹⁺¹ ÷ (−1 + 1) — this divides by zero; 1 ÷ x is the one power that integrates to a logarithm instead, ln|x| + C.
- Reaching for the power rule on a rational function whose denominator does not simplify away — check first whether partial fractions (denominator factors) or substitution (numerator matches part of the denominator's derivative) applies instead.
- Reporting a definite integral's numeric value as "the area" without checking sign — where the curve dips below the x-axis inside [a, b], that portion subtracts from the total, so a definite integral is signed area, not plain, unsigned area unless the curve stays on one side throughout.
Speed technique
- State the maxima/minima rule by shape every time: "curves down at the top, negative second derivative; curves up at the bottom, positive second derivative" — never memorize the sign pairing on its own.
- Before choosing an integration technique, scan the integrand: a product of two unrelated functions signals integration by parts, an inner function whose derivative also appears signals substitution, and a rational function with a factorable denominator signals partial fractions.
- For a tangent-and-normal question, compute f(x₀) and f'(x₀) once, then reuse both numbers for both lines — the tangent uses the slope directly, the normal uses its negative reciprocal.
- Check any antiderivative instantly by differentiating the answer back — it must reproduce the original integrand exactly, catching a sign error before it costs marks.
- For a 0 ÷ 0 limit, factor first and cancel the common factor — direct substitution after cancelling gives the limit in one line.
Check yourself
- Differentiate f(x) = x⁴ − 4x³ + 6 and evaluate f'(x) at x = 2.
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f'(2) = −16, since f'(x) = 4x³ − 12x², and 4(8) − 12(4) = 32 − 48 = −16. - Find the equation of the tangent to f(x) = x² at x = 3.
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y = 6x − 9, since f(3) = 9, f'(x) = 2x so f'(3) = 6, giving y − 9 = 6(x − 3). - For f(x) = −x² + 8x − 10, find the stationary point and state whether it is a maximum or minimum.
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x = 4, a maximum, since f'(x) = −2x + 8 = 0 gives x = 4, and f''(x) = −2 < 0 confirms a maximum, with f(4) = 6. - Evaluate ∫[0 to 2] 3x² dx using the Fundamental Theorem of Calculus.
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8, since an antiderivative is x³, and F(2) − F(0) = 8 − 0 = 8. - Decompose 1 ÷ [(x + 1)(x + 3)] into partial fractions.
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1 ÷ [2(x + 1)] − 1 ÷ [2(x + 3)], since 1 = A(x + 3) + B(x + 1) gives A = 1 ÷ 2 at x = −1 and B = −1 ÷ 2 at x = −3.
What the exam tests here
None of the 3 papers we hold has asked this. It is on the syllabus, so it can appear — but nothing in the paper record tells us how it would be framed. Treat it as insurance, not as a priority.
Try it: Calculus: Limits, Differentiation & Integration questions
Real questions from the CUET PG CS bank on exactly this skill. Pick an answer to see the full solution — the intuition, the worked steps, the faster methods and the traps.
Evaluate the limit: lim(x->0) .
Show the answer and worked solution
Answer: option C
Substituting x = 0 directly gives , so the expression needs simplifying first.
Factoring x out of the numerator gives x(x + 5)/x, and the common factor of x cancels for every x not equal to 0.
This leaves the limit of x + 5 as x approaches 0.
Substituting x = 0 into x + 5 gives 5, option C.
Evaluate the limit: lim(x->4) .
Show the answer and worked solution
Answer: option C
Direct substitution of x = 4 gives , so the numerator must be factored first.
- 16 factors as (x - 4)(x + 4), and the (x - 4) factor cancels with the denominator.
This leaves the limit of x + 4 as x approaches 4.
Substituting x = 4 into x + 4 gives 8, option C.
Evaluate the limit: lim(x->infinity) .
Show the answer and worked solution
Answer: option A
When the numerator and denominator of a rational function have the same highest power of x, the limit at infinity equals the ratio of their leading coefficients.
Here the highest power in both numerator and denominator is , with leading coefficients 3 and 1.
The limit is therefore .
This equals 3, option A.
Integration by parts, applied twice, gives the antiderivative of .e^(-x) as F(x) = −( + 2x + 2) (taking the constant of integration as 0). Find F(0).
Show the answer and worked solution
Answer: option A
Integration by parts on .e^(-x) needs to be applied a second time, to the leftover x.e^(-x) term from the first pass, producing the combined antiderivative F(x) = −(+2x+2).
Substituting x=0 into F(x) gives F(0) = -[+2(0)+2].
Since =1, this simplifies to −(2)(1).
The answer is −2, option A.
Using substitution, evaluate the definite integral: the integral of x( + 2)^3 dx from x = 0 to x = 2.
Show the answer and worked solution
Answer: option A
Let u = + 2, so du = 2x dx, meaning x dx = du/2.
The limits change too: at x=0, u=2, and at x=2, u=6, giving () times the integral of du from u=2 to u=6, with antiderivative ()() = .
Evaluating gives - = .
The answer is 160, option A.
Answer above — every one shows its working.