Math · Calculus

Limits and Derivatives revision notes for JEE

A concise revision note for Limits and Derivatives — the whole chapter on one page.

Revision note32 formulas alongside

01Key Concepts & Definitions

Calculus
Mathematics explaining the course of nature, dealing with the study of change in a function's value as points in its domain change.
Derivative
The instantaneous rate of change of a function, representing the slope of the tangent line to the function's curve at a specific point.
Limit
The expected value ll that a function f(x)f(x) approaches as the independent variable xx approaches a given value aa.
First Principle
The foundational definition of differentiation evaluating limits of the difference quotient as the interval approaches zero.

02Introduction & Intuitive Idea of Limits

Calculus is the branch of mathematics that mainly deals with the study of change in the value of a function as the points in the domain change. The concept of derivatives stems from the need to find the instantaneous rate of change, such as finding the exact velocity of a falling body at a specific second rather than an average velocity over an interval.

The instantaneous velocity v(t)v(t) at time t=at=a is equal to the slope of the tangent to the distance-time curve at t=at=a. To precisely define this slope, the mathematical concept of a "limit" is required.

03Left Hand and Right Hand Limits

The limit of a function f(x)f(x) as xx approaches aa is the expected value of f(x)f(x) based on the values of f(x)f(x) for points near aa.

  • Left Hand Limit (LHL): The expected value of f(x)f(x) at x=ax=a given the values of f(x)f(x) near xx to the left of aa (i.e., x<ax < a). It is denoted as lim⁡x→a−f(x)\lim_{x \to a^-} f(x).
  • Right Hand Limit (RHL): The expected value of f(x)f(x) at x=ax=a given the values of f(x)f(x) near xx to the right of aa (i.e., x>ax > a). It is denoted as lim⁡x→a+f(x)\lim_{x \to a^+} f(x).
  • Existence of Limit: If the right and left hand limits coincide, their common value is called the limit of f(x)f(x) at x=ax = a and is denoted by lim⁡x→af(x)\lim_{x \to a} f(x). If they are different, the limit does not exist, even if the function is defined at that point. JEE Tip Always check LHL and RHL separately for piecewise functions, modulus functions, and greatest integer functions.

04Algebra of Limits

Let ff and gg be two functions such that both lim⁡x→af(x)\lim_{x \to a} f(x) and lim⁡x→ag(x)\lim_{x \to a} g(x) exist.

  1. Sum Rule: lim⁡x→a[f(x)+g(x)]=lim⁡x→af(x)+lim⁡x→ag(x)\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x).
  2. Difference Rule: lim⁡x→a[f(x)−g(x)]=lim⁡x→af(x)−lim⁡x→ag(x)\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x).
  3. Product Rule: lim⁡x→a[f(x)⋅g(x)]=lim⁡x→af(x)⋅lim⁡x→ag(x)\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x).
  4. Quotient Rule: lim⁡x→a[f(x)g(x)]=lim⁡x→af(x)lim⁡x→ag(x)\lim_{x \to a} \left[\frac{f(x)}{g(x)}\right] = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, provided lim⁡x→ag(x)≠0\lim_{x \to a} g(x) \neq 0.
  5. Scalar Multiple Rule: lim⁡x→a[λ⋅f(x)]=λ⋅lim⁡x→af(x)\lim_{x \to a} [\lambda \cdot f(x)] = \lambda \cdot \lim_{x \to a} f(x) for any real number λ\lambda.

05Limits of Polynomials and Rational Functions

A polynomial function is of the form f(x)=a0+a1x+a2x2+⋯+anxnf(x) = a_0 + a_1x + a_2x^2 + \dots + a_nx^n.

  • The limit of a polynomial function at x=ax = a is simply the value of the function at aa: lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).

A rational function is f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}, where g(x)g(x) and h(x)h(x) are polynomials.

  • If h(a)≠0h(a) \neq 0, then lim⁡x→af(x)=g(a)h(a)\lim_{x \to a} f(x) = \frac{g(a)}{h(a)}.
  • If h(a)=0h(a) = 0 and g(a)≠0g(a) \neq 0, the limit does not exist.
  • If h(a)=0h(a) = 0 and g(a)=0g(a) = 0, this is a 0/00/0 indeterminate form. To evaluate, factorize g(x)g(x) and h(x)h(x) as g(x)=(x−a)kg1(x)g(x) = (x-a)^k g_1(x) and h(x)=(x−a)lh1(x)h(x) = (x-a)^l h_1(x), cancel the common (x−a)(x-a) terms, and then evaluate. JEE Tip We can safely cancel (x−a)(x-a) from numerator and denominator because the limit operation implies x→ax \to a, which strictly means x≠ax \neq a.

Standard Limit Formula for Powers

For any positive integer nn (and by extension, any rational number), lim⁡x→axn−anx−a=nan−1\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}. JEE Tip This is highly useful for fractional powers where binomial expansion is tedious.

06Limits of Trigonometric Functions

Limits of trigonometric functions rely on the order properties of these functions:

  • Theorem: If f(x)≤g(x)f(x) \leq g(x) for all xx in the domain, and their limits exist at aa, then lim⁡x→af(x)≤lim⁡x→ag(x)\lim_{x \to a} f(x) \leq \lim_{x \to a} g(x).
  • Sandwich Theorem (Squeeze Theorem): Let f,gf, g, and hh be real functions such that f(x)≤g(x)≤h(x)f(x) \leq g(x) \leq h(x) for all xx in the common domain. If lim⁡x→af(x)=l=lim⁡x→ah(x)\lim_{x \to a} f(x) = l = \lim_{x \to a} h(x), then lim⁡x→ag(x)=l\lim_{x \to a} g(x) = l. JEE Tip The Sandwich Theorem is the primary tool for solving limits involving oscillating functions like sin⁡(1/x)\sin(1/x) combined with polynomials, e.g., lim⁡x→0x2sin⁡(1/x)=0\lim_{x \to 0} x^2 \sin(1/x) = 0.

Fundamental Trigonometric Inequality

For 0<x<π20 < x < \frac{\pi}{2} (where xx is in radians), sin⁡x<x<tan⁡x\sin x < x < \tan x.

Standard Trigonometric Limits

  1. lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1.
  2. lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{1 - \cos x}{x} = 0.
  3. lim⁡x→0tan⁡xx=1\lim_{x \to 0} \frac{\tan x}{x} = 1. JEE Tip These limits are valid ONLY when xx is measured in RADIANS. If xx is in degrees, lim⁡x→0sin⁡x∘x=π180\lim_{x \to 0} \frac{\sin x^\circ}{x} = \frac{\pi}{180}.

07Derivatives & The First Principle

The derivative of a function quantifies the rate of change of f(x)f(x) with respect to xx. Geometrically, the derivative of f(x)f(x) at x=ax=a is the slope of the tangent to the curve y=f(x)y=f(x) at the point (a,f(a))(a, f(a)).

Derivative from First Principle

The derivative of ff at xx is defined as: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. It is denoted by f′(x)f'(x), ddx[f(x)]\frac{d}{dx}[f(x)], dydx\frac{dy}{dx}, or D(f(x))D(f(x)).

08Algebra of Derivatives

Let f(x)=uf(x) = u and g(x)=vg(x) = v be functions whose derivatives exist.

  1. Sum Rule: (u+v)′=u′+v′(u + v)' = u' + v'.
  2. Difference Rule: (u−v)′=u′−v′(u - v)' = u' - v'.
  3. Product Rule (Leibnitz Rule): (uv)′=u′v+uv′(uv)' = u'v + uv'.
  4. Quotient Rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} (where v≠0v \neq 0). JEE Tip Trap alert! The negative sign in the numerator is specifically in front of the term uv′uv', which differentiates the denominator. Order matters!

09Standard Derivatives

  1. Constant Function: ddx(a)=0\frac{d}{dx}(a) = 0.
  2. Identity Function: ddx(x)=1\frac{d}{dx}(x) = 1.
  3. Power Rule: ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1} for any real number nn.
  4. Polynomial Function: ddx(anxn+an−1xn−1+⋯+a0)=nanxn−1+(n−1)an−1xn−2+⋯+a1\frac{d}{dx}(a_n x^n + a_{n-1} x^{n-1} + \dots + a_0) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1.
  5. Trigonometric Functions:
    • ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x.
    • ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x.
    • ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x.
    • ddx(cot⁡x)=−cosec2x\frac{d}{dx}(\cot x) = -\text{cosec}^2 x.
    • ddx(sec⁡x)=sec⁡xtan⁡x\frac{d}{dx}(\sec x) = \sec x \tan x [Derived using quotient rule on 1/cos⁡x1/\cos x].
    • ddx(cosecx)=−cosecxcot⁡x\frac{d}{dx}(\text{cosec} x) = -\text{cosec} x \cot x.

10JEE Advanced Extensions (Added Topics)

  • L'Hôpital's Rule: If lim⁡x→af(x)g(x)\lim_{x \to a} \frac{f(x)}{g(x)} results in 0/00/0 or ∞/∞\infty/\infty, then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, provided the latter limit exists. JEE Tip Always check if the form is actually 0/00/0 or ∞/∞\infty/\infty before applying L'Hôpital's Rule. Applying it to finite forms gives incorrect answers!
  • 1∞1^\infty Form Limits: If lim⁡x→af(x)=1\lim_{x \to a} f(x) = 1 and lim⁡x→ag(x)=∞\lim_{x \to a} g(x) = \infty, then lim⁡x→af(x)g(x)=elim⁡x→ag(x)[f(x)−1]\lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x)[f(x) - 1]}.
Standard Maclaurin Series Expansions

Extremely powerful for complex 0/00/0 limits:

sin⁡x=x−x33!+x55!−…\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots

cos⁡x=1−x22!+x44!−…\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots

tan⁡x=x+x33+2x515+…\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots

ex=1+x+x22!+x33!+…e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots

ln⁡(1+x)=x−x22+x33−…\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots

  • Chain Rule: ddx[f(g(x))]=f′(g(x))⋅g′(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x).

11Formulae, Equations & Units

  • Average Velocity: vavg=ΔsΔt=s(t2)−s(t1)t2−t1v_{avg} = \frac{\Delta s}{\Delta t} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}. Unit: m/s (meters per second).
  • Instantaneous Rate of Change / Limit definition of Derivative: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
  • Standard Algebraic Limit: lim⁡x→axn−anx−a=nan−1\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}.
  • Half-Angle identity limit evaluation: 1−cos⁡x=2sin⁡2(x2)1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) is frequently used to prove lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{1-\cos x}{x} = 0.

12Conditions & Limitations

  • Limit Existence Constraint: lim⁡x→af(x)\lim_{x \to a} f(x) ONLY exists if the Left Hand Limit equals the Right Hand Limit.
  • Quotient Rule Constraint: ddx(uv)\frac{d}{dx}\left(\frac{u}{v}\right) and limits of quotients lim⁡f(x)g(x)\lim \frac{f(x)}{g(x)} are valid strictly where the denominator v≠0v \neq 0 or lim⁡g(x)≠0\lim g(x) \neq 0 respectively.
  • Rational Limit Division Constraint: When canceling factors like (x−a)(x-a) out of the form 0/00/0, the mathematical assumption validating the cancellation is x→ax \to a implies x≠ax \neq a (hence you are not dividing by exactly zero).

13COMMON MISCONCEPTIONS & SIGN CONVENTIONS

  • Misconception: f(a)f(a) is always equal to lim⁡x→af(x)\lim_{x \to a} f(x). Fact: The limit depends strictly on the values around x=ax=a, not the value at x=ax=a. The value f(a)f(a) and the limit can be completely different, or one may exist while the other does not.
  • Misconception: Infinite limit means limit exists. Fact: If lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \infty, the limit technically does not exist because infinity is not a real number.
  • Misconception: Radians and degrees are interchangeable in limits. Fact: The formulas lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 and derivative ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x STRICTLY assume xx is measured in radians. If xx is in degrees, the limit is π/180\pi/180.
  • Sign Convention in Quotient Rule: Always remember it is (Numerator′×Denominator)−(Numerator×Denominator′)(\text{Numerator}' \times \text{Denominator}) - (\text{Numerator} \times \text{Denominator}'), divided by Denominator2\text{Denominator}^2. Reversing the negative sign is a fatal algebraic error.

14Previous Year JEE Topics

  • 1∞1^\infty limits: Used in virtually every JEE paper.
  • Limits with Greatest Integer Functions [x][x] and Fractional Parts {x}\{x\}: These require breaking limits carefully into LHL and RHL because these functions jump at integer points.
  • Evaluating Limits using Taylor/Maclaurin Series Expansion: For complex polynomial/trigonometric mixed fractions where L'Hôpital's rule becomes too lengthy.
  • Derivatives of Implicit Functions & Inverse Trigonometric Functions.
  • Checking continuity and differentiability via First Principles.

15Standard Derivations & Step-by-Step Problem Solving

1. Derivation of lim⁡x→axn−anx−a=nan−1\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} (for integer nn)

  • Method 1 (Factorization): Divide (xn−an)(x^n - a^n) by (x−a)(x - a). xn−an=(x−a)(xn−1+xn−2a+xn−3a2+⋯+xan−2+an−1)x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + x a^{n-2} + a^{n-1}). Taking the limit as x→ax \to a of the second bracket: lim⁡x→a(xn−1+axn−2+⋯+an−1)=an−1+a(an−2)+⋯+an−1\lim_{x \to a} (x^{n-1} + a x^{n-2} + \dots + a^{n-1}) = a^{n-1} + a(a^{n-2}) + \dots + a^{n-1}. There are nn terms, so the sum is nan−1n a^{n-1}.
  • Method 2 (Binomial Theorem for x→ax \to a or h→0h \to 0): Let x=a+hx = a + h. Then lim⁡h→0(a+h)n−anh\lim_{h \to 0} \frac{(a+h)^n - a^n}{h}. Expand (a+h)n(a+h)^n using binomial theorem, the ana^n cancels, divide by hh, and substituting h=0h=0 leaves only nan−1n a^{n-1}.

2. Geometric Proof of sin⁡x<x<tan⁡x\sin x < x < \tan x

  • Consider a unit circle with center OO and an angle xx (in radians) such that 0<x<π/20 < x < \pi/2.
  • Area of ΔOAC<Area of Sector OAC<Area of ΔOAB\Delta OAC < \text{Area of Sector } OAC < \text{Area of } \Delta OAB.
  • Area ΔOAC=12OA⋅CD=12(1)sin⁡x=12sin⁡x\Delta OAC = \frac{1}{2} OA \cdot CD = \frac{1}{2} (1) \sin x = \frac{1}{2} \sin x.
  • Area of Sector OAC=12r2x=12(1)2x=12xOAC = \frac{1}{2} r^2 x = \frac{1}{2} (1)^2 x = \frac{1}{2} x.
  • Area of ΔOAB=12OA⋅AB=12(1)tan⁡x=12tan⁡x\Delta OAB = \frac{1}{2} OA \cdot AB = \frac{1}{2} (1) \tan x = \frac{1}{2} \tan x.
  • Therefore, sin⁡x<x<tan⁡x\sin x < x < \tan x. Dividing by sin⁡x\sin x gives 1<xsin⁡x<1cos⁡x1 < \frac{x}{\sin x} < \frac{1}{\cos x}, taking reciprocals gives cos⁡x<sin⁡xx<1\cos x < \frac{\sin x}{x} < 1, which proves lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 via Sandwich Theorem.

3. Derivative of sin⁡x\sin x from First Principle

f′(x)=lim⁡h→0sin⁡(x+h)−sin⁡xhf'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}. Using the formula sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right): f′(x)=lim⁡h→02cos⁡(x+h2)sin⁡(h2)hf'(x) = \lim_{h \to 0} \frac{2 \cos(x + \frac{h}{2}) \sin(\frac{h}{2})}{h}. =lim⁡h→0cos⁡(x+h2)⋅lim⁡h→0sin⁡(h/2)h/2=cos⁡(x)⋅1=cos⁡x= \lim_{h \to 0} \cos(x + \frac{h}{2}) \cdot \lim_{h \to 0} \frac{\sin(h/2)}{h/2} = \cos(x) \cdot 1 = \cos x.

16JEE Traps

Phantom Zeroes
✕Misconception

Direct substitution in 0/00/0 forms gives 00 or ∞\infty.

✓Reality

0/00/0 is an indeterminate form. You must cancel the vanishing factor (e.g., (x−a)(x-a)) from the numerator and denominator, apply L'Hôpital's rule, or use series expansions.

Coefficient Missing
✕Misconception

lim⁡x→0sin⁡(kx)x=1\lim_{x \to 0} \frac{\sin(kx)}{x} = 1.

✓Reality

lim⁡x→0sin⁡(kx)x=k\lim_{x \to 0} \frac{\sin(kx)}{x} = k. You must multiply and divide by kk to apply the standard identity lim⁡θ→0sin⁡θθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1.

Degree Trap
✕Misconception

Trigonometric limit identities work in degrees.

✓Reality

lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 is ONLY valid when xx is in radians. For xx in degrees, the limit is π180\frac{\pi}{180}.

GIF Discontinuity
✕Misconception

lim⁡x→0[x]=0\lim_{x \to 0} [x] = 0 (where [x][x] is the greatest integer function).

✓Reality

The limit does NOT exist. LHL = lim⁡x→0−[x]=−1\lim_{x \to 0^-} [x] = -1, while RHL = lim⁡x→0+[x]=0\lim_{x \to 0^+} [x] = 0.

Modulus Trap
✕Misconception

When taking limits to infinity involving x2\sqrt{x^2}, you can just write xx.

✓Reality

x2=∣x∣\sqrt{x^2} = |x|. If x→−∞x \to -\infty, then x2=−x\sqrt{x^2} = -x. Missing this minus sign ruins the whole problem.

Local Neighborhoods
✕Misconception

Sandwich theorem applies if f(x)≤g(x)≤h(x)f(x) \leq g(x) \leq h(x) at exactly one point.

✓Reality

The inequality must hold for an entire neighborhood around the limit point (except possibly at the point itself).

Sign Error in Division
✕Misconception

In the quotient rule, (u/v)′=u′v+uv′v2(u/v)' = \frac{u'v + uv'}{v^2}.

✓Reality

The sign is negative: u′v−uv′v2\frac{u'v - uv'}{v^2}. Always start with the derivative of the numerator.

Combining DNE limits
✕Misconception

If lim⁡f(x)\lim f(x) does not exist and lim⁡g(x)\lim g(x) does not exist, then lim⁡[f(x)+g(x)]\lim [f(x) + g(x)] does not exist.

✓Reality

The sum can easily have a limit. Example: f(x)=[x]f(x) = [x] and g(x)=−[x]g(x) = -[x] at integers.

Sharp Corners
✕Misconception

The derivative of a function at x=ax=a is always defined if the function is continuous.

✓Reality

Continuous functions can have sharp corners (like ∣x∣|x| at x=0x=0) where LHD ≠\neq RHD, meaning the derivative does not exist.

Squared Denominator in Cosine Limits
✕Misconception

lim⁡x→01−cos⁡xx2=0\lim_{x \to 0} \frac{1-\cos x}{x^2} = 0.

✓Reality

lim⁡x→01−cos⁡xx=0\lim_{x \to 0} \frac{1-\cos x}{x} = 0, but lim⁡x→01−cos⁡xx2=12\lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2} (derived using the half-angle formula 1−cos⁡x=2sin⁡2(x/2)1-\cos x = 2\sin^2(x/2)).

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