Math · Calculus

Limits and Derivatives revision notes

A concise JEE revision summary of Limits and Derivatives.

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Mathrevision notes

Introduction & 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.

Left 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 limxaf(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 limxa+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 limxaf(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.

Algebra of Limits

Let ff and gg be two functions such that both limxaf(x)\lim_{x \to a} f(x) and limxag(x)\lim_{x \to a} g(x) exist.

  1. Sum Rule: limxa[f(x)+g(x)]=limxaf(x)+limxag(x)\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x).
  2. Difference Rule: limxa[f(x)g(x)]=limxaf(x)limxag(x)\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x).
  3. Product Rule: limxa[f(x)g(x)]=limxaf(x)limxag(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: limxa[f(x)g(x)]=limxaf(x)limxag(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 limxag(x)0\lim_{x \to a} g(x) \neq 0.
  5. Scalar Multiple Rule: limxa[λf(x)]=λlimxaf(x)\lim_{x \to a} [\lambda \cdot f(x)] = \lambda \cdot \lim_{x \to a} f(x) for any real number λ\lambda.

Limits 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: limxaf(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 limxaf(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)=(xa)kg1(x)g(x) = (x-a)^k g_1(x) and h(x)=(xa)lh1(x)h(x) = (x-a)^l h_1(x), cancel the common (xa)(x-a) terms, and then evaluate. → [JEE TIP] We can safely cancel (xa)(x-a) from numerator and denominator because the limit operation implies xax \to a, which strictly means xax \neq a.

Standard Limit Formula for Powers: For any positive integer nn (and by extension, any rational number), limxaxnanxa=nan1\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.

Limits 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 limxaf(x)limxag(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 limxaf(x)=l=limxah(x)\lim_{x \to a} f(x) = l = \lim_{x \to a} h(x), then limxag(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., limx0x2sin(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), sinx<x<tanx\sin x < x < \tan x.

Standard Trigonometric Limits:

  1. limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1.
  2. limx01cosxx=0\lim_{x \to 0} \frac{1 - \cos x}{x} = 0.
  3. limx0tanxx=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, limx0sinxx=π180\lim_{x \to 0} \frac{\sin x^\circ}{x} = \frac{\pi}{180}.

Derivatives & 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)=limh0f(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)).

Algebra 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: (uv)=uv(u - v)' = u' - v'.
  3. Product Rule (Leibnitz Rule): (uv)=uv+uv(uv)' = u'v + uv'.
  4. Quotient Rule: (uv)=uvuvv2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} (where v0v \neq 0). → [JEE TIP] Trap alert! The negative sign in the numerator is specifically in front of the term uvuv', which differentiates the denominator. Order matters!

Standard 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)=nxn1\frac{d}{dx}(x^n) = nx^{n-1} for any real number nn.
  4. Polynomial Function: ddx(anxn+an1xn1++a0)=nanxn1+(n1)an1xn2++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(sinx)=cosx\frac{d}{dx}(\sin x) = \cos x.
    • ddx(cosx)=sinx\frac{d}{dx}(\cos x) = -\sin x.
    • ddx(tanx)=sec2x\frac{d}{dx}(\tan x) = \sec^2 x.
    • ddx(cotx)=cosec2x\frac{d}{dx}(\cot x) = -\text{cosec}^2 x.
    • ddx(secx)=secxtanx\frac{d}{dx}(\sec x) = \sec x \tan x [Derived using quotient rule on 1/cosx1/\cos x].
    • ddx(cosecx)=cosecxcotx\frac{d}{dx}(\text{cosec} x) = -\text{cosec} x \cot x.

JEE Advanced Extensions (Added Topics)

  • L'Hôpital's Rule: If limxaf(x)g(x)\lim_{x \to a} \frac{f(x)}{g(x)} results in 0/00/0 or /\infty/\infty, then limxaf(x)g(x)=limxaf(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!
  • 11^\infty Form Limits: If limxaf(x)=1\lim_{x \to a} f(x) = 1 and limxag(x)=\lim_{x \to a} g(x) = \infty, then limxaf(x)g(x)=elimxag(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:
    • sinx=xx33!+x55!\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots
    • cosx=1x22!+x44!\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots
    • tanx=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)=xx22+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).

Key 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 lll that a function f(x)f(x)f(x) approaches as the independent variable xxx approaches a given value aaa.
First Principle:
The foundational definition of differentiation evaluating limits of the difference quotient as the interval approaches zero.

Formulae, Equations & Units

  • Average Velocity: vavg=ΔsΔt=s(t2)s(t1)t2t1v_{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)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
  • Standard Algebraic Limit: limxaxnanxa=nan1\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}.
  • Half-Angle identity limit evaluation: 1cosx=2sin2(x2)1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) is frequently used to prove limx01cosxx=0\lim_{x \to 0} \frac{1-\cos x}{x} = 0.

Conditions & Limitations

  • Limit Existence Constraint: limxaf(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 limf(x)g(x)\lim \frac{f(x)}{g(x)} are valid strictly where the denominator v0v \neq 0 or limg(x)0\lim g(x) \neq 0 respectively.
  • Rational Limit Division Constraint: When canceling factors like (xa)(x-a) out of the form 0/00/0, the mathematical assumption validating the cancellation is xax \to a implies xax \neq a (hence you are not dividing by exactly zero).

⚠️ COMMON MISCONCEPTIONS & SIGN CONVENTIONS

  • Misconception: f(a)f(a) is always equal to limxaf(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 limxaf(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 limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 and derivative ddx(sinx)=cosx\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.

Previous Year JEE Topics

  • 11^\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.

Standard Derivations & Step-by-Step Problem Solving

1. Derivation of limxaxnanxa=nan1\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} (for integer nn):

  • Method 1 (Factorization): Divide (xnan)(x^n - a^n) by (xa)(x - a). xnan=(xa)(xn1+xn2a+xn3a2++xan2+an1)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 xax \to a of the second bracket: limxa(xn1+axn2++an1)=an1+a(an2)++an1\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 nan1n a^{n-1}.
  • Method 2 (Binomial Theorem for xax \to a or h0h \to 0): Let x=a+hx = a + h. Then limh0(a+h)nanh\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 nan1n a^{n-1}.

2. Geometric Proof of sinx<x<tanx\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=12OACD=12(1)sinx=12sinx\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=12OAAB=12(1)tanx=12tanx\Delta OAB = \frac{1}{2} OA \cdot AB = \frac{1}{2} (1) \tan x = \frac{1}{2} \tan x.
  • Therefore, sinx<x<tanx\sin x < x < \tan x. Dividing by sinx\sin x gives 1<xsinx<1cosx1 < \frac{x}{\sin x} < \frac{1}{\cos x}, taking reciprocals gives cosx<sinxx<1\cos x < \frac{\sin x}{x} < 1, which proves limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1 via Sandwich Theorem.

3. Derivative of sinx\sin x from First Principle: f(x)=limh0sin(x+h)sinxhf'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}. Using the formula sinAsinB=2cos(A+B2)sin(AB2)\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right): f(x)=limh02cos(x+h2)sin(h2)hf'(x) = \lim_{h \to 0} \frac{2 \cos(x + \frac{h}{2}) \sin(\frac{h}{2})}{h}. =limh0cos(x+h2)limh0sin(h/2)h/2=cos(x)1=cosx= \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.

Memory Aids & JEE Traps

Based on this chapter, here are the top 10 MCQ traps and tricks that appear in JEE:

  1. Misconception → Direct substitution in 0/00/0 forms gives 00 or \infty. Correct Understanding0/00/0 is an indeterminate form. You must cancel the vanishing factor (e.g., (xa)(x-a)) from the numerator and denominator, apply L'Hôpital's rule, or use series expansions. → [JEE TIP] Trap 1 - Phantom Zeroes.
  2. Misconceptionlimx0sin(kx)x=1\lim_{x \to 0} \frac{\sin(kx)}{x} = 1. Correct Understandinglimx0sin(kx)x=k\lim_{x \to 0} \frac{\sin(kx)}{x} = k. You must multiply and divide by kk to apply the standard identity limθ0sinθθ=1\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1. → [JEE TIP] Trap 2 - Coefficient Missing.
  3. Misconception → Trigonometric limit identities work in degrees. Correct Understandinglimx0sinxx=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}. → [JEE TIP] Trap 3 - Degree Trap.
  4. Misconceptionlimx0[x]=0\lim_{x \to 0} [x] = 0 (where [x][x] is the greatest integer function). Correct Understanding → The limit does NOT exist. LHL = limx0[x]=1\lim_{x \to 0^-} [x] = -1, while RHL = limx0+[x]=0\lim_{x \to 0^+} [x] = 0. → [JEE TIP] Trap 4 - GIF Discontinuity.
  5. Misconception → When taking limits to infinity involving x2\sqrt{x^2}, you can just write xx. Correct Understandingx2=x\sqrt{x^2} = |x|. If xx \to -\infty, then x2=x\sqrt{x^2} = -x. Missing this minus sign ruins the whole problem. → [JEE TIP] Trap 5 - Modulus Trap.
  6. Misconception → Sandwich theorem applies if f(x)g(x)h(x)f(x) \leq g(x) \leq h(x) at exactly one point. Correct Understanding → The inequality must hold for an entire neighborhood around the limit point (except possibly at the point itself). → [JEE TIP] Trap 6 - Local Neighborhoods.
  7. Misconception → In the quotient rule, (u/v)=uv+uvv2(u/v)' = \frac{u'v + uv'}{v^2}. Correct Understanding → The sign is negative: uvuvv2\frac{u'v - uv'}{v^2}. Always start with the derivative of the numerator. → [JEE TIP] Trap 7 - Sign Error in Division.
  8. Misconception → If limf(x)\lim f(x) does not exist and limg(x)\lim g(x) does not exist, then lim[f(x)+g(x)]\lim [f(x) + g(x)] does not exist. Correct Understanding → The sum can easily have a limit. Example: f(x)=[x]f(x) = [x] and g(x)=[x]g(x) = -[x] at integers. → [JEE TIP] Trap 8 - Combining DNE limits.
  9. Misconception → The derivative of a function at x=ax=a is always defined if the function is continuous. Correct Understanding → Continuous functions can have sharp corners (like x|x| at x=0x=0) where LHD \neq RHD, meaning the derivative does not exist. → [JEE TIP] Trap 9 - Sharp Corners.
  10. Misconceptionlimx01cosxx2=0\lim_{x \to 0} \frac{1-\cos x}{x^2} = 0. Correct Understandinglimx01cosxx=0\lim_{x \to 0} \frac{1-\cos x}{x} = 0, but limx01cosxx2=12\lim_{x \to 0} \frac{1-\cos x}{x^2} = \frac{1}{2} (derived using the half-angle formula 1cosx=2sin2(x/2)1-\cos x = 2\sin^2(x/2)). → [JEE TIP] Trap 10 - Squared Denominator in Cosine Limits.
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