1. Chapter at a Glance
Calculus begins with the idea of a limit — the value a function approaches as x gets close to a point. Continuity means the graph has no break, and differentiability means the graph is smooth, with a well-defined tangent. This chapter also covers the methods of differentiation. Together they give 2–3 questions per JEE Main shift, so they are worth mastering thoroughly.
- Evaluating limits of the forms 0/0, ∞/∞ and 1∞ with standard limits or L'Hospital's rule.
- Finding a constant that makes a piecewise function continuous or differentiable.
- Counting points of non-differentiability of functions with |x| or [x].
- Derivatives of implicit, parametric and xx-type functions.
2. Limits
We write limx→a f(x) = L if f(x) gets as close as we like to L when x is close enough to a (x ≠ a). The limit exists only when the left-hand limit (LHL) and the right-hand limit (RHL) exist and are equal.
Standard limits
| Limit (x → 0 unless stated) | Value |
|---|---|
| sin x / x, tan x / x | 1 |
| (1 − cos x)/x² | 1/2 |
| (eˣ − 1)/x | 1 |
| (aˣ − 1)/x | ln a |
| ln(1 + x)/x | 1 |
| (1 + x)1/x | e |
| (xⁿ − aⁿ)/(x − a), x → a | naⁿ⁻¹ |
| sin−1x / x, tan−1x / x | 1 |
The 1∞ form
L'Hospital's rule
For the forms 0/0 or ∞/∞: lim f(x)/g(x) = lim f′(x)/g′(x), provided the right side exists. Apply repeatedly if the form persists.
Series expansions (very useful)
- eˣ = 1 + x + x²/2! + x³/3! + …
- sin x = x − x³/3! + x⁵/5! − …, cos x = 1 − x²/2! + x⁴/4! − …
- tan x = x + x³/3 + 2x⁵/15 + …, ln(1 + x) = x − x²/2 + x³/3 − …
Limits at infinity
For a ratio of polynomials, divide by the highest power of x in the denominator: the limit is the ratio of leading coefficients (equal degrees), 0 (numerator of lower degree) or ±∞ (numerator of higher degree).
3. Continuity
f is continuous at x = a if LHL = RHL = f(a). Informally, you can draw the graph near a without lifting the pen.
- Polynomials, eˣ, sin x, cos x and |x| are continuous everywhere; log x is continuous on x > 0.
- Sums, differences, products and compositions of continuous functions are continuous; quotients are continuous wherever the denominator is non-zero.
- [x] is discontinuous at every integer; {x} likewise.
- Intermediate Value Theorem: if f is continuous on [a, b] and k lies between f(a) and f(b), then f(c) = k for some c in [a, b]. In particular, a sign change means a root exists.
4. Differentiability
f is differentiable at a if the left-hand derivative (LHD) and right-hand derivative (RHD) are equal:
Differentiable ⇒ continuous, but continuous does not imply differentiable (|x| at 0). So if a function is discontinuous at a point, it is certainly not differentiable there.
- |f(x)| may fail to be differentiable where f(x) = 0 and f changes sign (a sharp corner).
- A sum of the form |x − a1| + |x − a2| + … is non-differentiable at each distinct ai.
- x|x| is differentiable everywhere (the corner is smoothed out).
5. Methods of Differentiation
| Function | Derivative |
|---|---|
| xⁿ | nxⁿ⁻¹ |
| eˣ, aˣ | eˣ, aˣ ln a |
| ln x | 1/x |
| sin x, cos x | cos x, −sin x |
| tan x, sec x | sec²x, sec x tan x |
| sin−1x | 1/√(1 − x²) |
| tan−1x | 1/(1 + x²) |
- Product rule: (uv)′ = u′v + uv′. Quotient rule: (u/v)′ = (u′v − uv′)/v².
- Chain rule: d/dx f(g(x)) = f′(g(x)) g′(x).
- Implicit differentiation: differentiate both sides with respect to x, treating y as a function of x.
- Logarithmic differentiation: for y = uv or long products, take log first. For y = xˣ: y′ = xˣ(1 + ln x).
- Parametric: dy/dx = (dy/dt)/(dx/dt); d²y/dx² = d/dt(dy/dx) ÷ (dx/dt).
- Substitution in inverse trig: e.g. y = tan−1(2x/(1 − x²)) = 2tan−1x for |x| < 1, so y′ = 2/(1 + x²).
6. Formula Master Table
| Tool | Formula |
|---|---|
| Limit exists | LHL = RHL |
| Continuity at a | LHL = RHL = f(a) |
| Derivative at a | lim [f(a + h) − f(a)]/h |
| 1∞ form | elim g(f − 1) |
| L'Hospital | lim f/g = lim f′/g′ |
| Parametric derivative | dy/dx = ẏ/ẋ |
| xˣ | xˣ(1 + ln x) |
| Derivative of inverse function | (f−1)′(y) = 1/f′(x) |
| Leibniz rule | d/dx ∫a(x)b(x) f(t)dt = f(b)b′ − f(a)a′ |
7. Mind Map
- LHL = RHL
- Standard limits
- L'Hospital, series
- 1∞ → e…
- LHL = RHL = f(a)
- Jumps, holes
- IVT
- LHD = RHD
- Corners of |x|
- Diff ⇒ cont
- Chain, product, quotient
- Implicit, parametric
- Log differentiation
8. Tips and Tricks
- For limits as x → 0, replace sin x, tan x, eˣ − 1, ln(1 + x) by x, and 1 − cos x by x²/2, when they appear as factors (not in a sum where the leading terms cancel).
- If the leading terms cancel (like tan x − sin x), use the series up to x³: tan x − sin x ≈ x³/2.
- For "find k so that f is continuous", simply equate the limit from both sides with the function value.
- For |f(x)|, look for simple roots of f (sign changes) — each is a likely corner.
- Applying L'Hospital's rule to a limit that is not of the form 0/0 or ∞/∞.
- Using the quotient rule instead of simplifying first — simplification often avoids long work.
- Assuming a function is differentiable just because it is continuous.
9. Important One-liners
- limx→∞ (1 + a/x)ˣ = eᵃ.
- The greatest integer function is continuous from the right at integers but not from the left.
- If f is differentiable at a, it is continuous at a.
- |x| is continuous everywhere but not differentiable at 0.
- The derivative of an even function is odd, and the derivative of an odd function is even.
- A polynomial is continuous and differentiable everywhere.
- If f(x) = |x − 1| + |x − 2| + |x − 3|, it is non-differentiable at exactly 3 points.
- d/dx (tan−1x + cot−1x) = 0 because their sum is the constant π/2.
10. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
limx→0 (sin 3x)/(tan 5x) equals:
- 5/3
- 3/5
- 1
- 15
Show solution
Write it as [sin 3x/(3x)] × [5x/tan 5x] × (3x/5x). The first two factors tend to 1, leaving 3/5.
Answer: (B)
Find limx→0 (1 − cos 2x)/x².
Show solution
1 − cos 2x = 2 sin²x, so the limit = 2(sin x/x)² → 2 × 1 = 2.
Answer: 2
limx→∞ (1 + 2/x)ˣ equals:
- e
- 2e
- e²
- 1
Show solution
It is of the form 1∞: limit = elim x·(2/x) = e².
Answer: (C)
limx→0 (tan x − sin x)/x³ equals:
- 0
- 1
- 1/2
- 1/6
Show solution
tan x − sin x = tan x (1 − cos x). So the limit = (tan x/x) × ((1 − cos x)/x²) → 1 × 1/2 = 1/2.
(Series check: tan x − sin x = (x + x³/3) − (x − x³/6) + … = x³/2 + ….)
Answer: (C)
f(x) = (x² − 9)/(x − 3) for x ≠ 3 and f(3) = k. Find k so that f is continuous at x = 3.
Show solution
For x ≠ 3, f(x) = x + 3, so limx→3 f(x) = 6. For continuity k = 6.
Answer: 6
Find the number of points at which f(x) = |x| + |x − 1| + |x − 2| is not differentiable.
Show solution
Each term |x − a| has a corner at x = a. The corners are at x = 0, 1 and 2, and they do not cancel because all terms have positive coefficients (the slope jumps by +2 at each).
Answer: 3
If x = a cos³t and y = a sin³t, then dy/dx equals:
- tan t
- −tan t
- −cot t
- cot t
Show solution
dx/dt = −3a cos²t sin t and dy/dt = 3a sin²t cos t.
dy/dx = (3a sin²t cos t)/(−3a cos²t sin t) = −sin t/cos t = −tan t.
Answer: (B)
If y = xˣ, find dy/dx at x = 1.
Show solution
ln y = x ln x ⇒ (1/y) y′ = ln x + 1 ⇒ y′ = xˣ(1 + ln x).
At x = 1: y′ = 1 × (1 + 0) = 1.
Answer: 1
limx→0 (eˣ − 1 − x)/x² equals:
- 1
- 1/2
- 0
- 2
Show solution
Using eˣ = 1 + x + x²/2 + …, the numerator is x²/2 + x³/6 + …, so the limit is 1/2.
(L'Hospital twice gives the same: eˣ/2 → 1/2.)
Answer: (B)
11. Quick Revision Summary
- Limit exists ⇔ LHL = RHL; continuity also needs = f(a).
- sin x/x → 1, (1 − cos x)/x² → 1/2, (eˣ − 1)/x → 1, 1∞ → elim g(f−1).
- Differentiable ⇒ continuous; corners of |x − a| break differentiability.
- Log differentiation for xˣ; parametric dy/dx = ẏ/ẋ.
- Use series expansions when the leading terms cancel.