Limits and Derivatives

Every real change — a car speeding up, a chemical reaction proceeding, a falling stone gathering speed — poses the same question: how fast is something changing right now, at one instant, not on average over some stretch of time? Ordinary algebra tells you how far a body travels in five seconds; it has no word for "at this instant." This chapter builds that word, and it does so by chaining together two ideas that turn out to be the same idea seen twice: the limit and the derivative. We begin exactly as NCERT does, with a stone dropped from a cliff, $s = 4.9t^2$ metres in $t$ seconds. Average velocity over any interval is easy — divide distance by time — but the average velocity from $t=1.99$ to $t=2$ is not the velocity at $t=2$ . To pin that down, the chapter shrinks the interval smaller and smaller and watches the average velocity settle toward a single number. That settling-down process is the seed of the idea of a limit: the value a function is heading toward as its input approaches a point, whether or not the function is even defined there. From this intuition the chapter builds a formal toolkit: the left hand limit and right hand limit, an algebra of limits that lets a complicated limit be broken into simpler pieces, a standard result for $\lim_{x\to a}\frac{x^n-a^n}{x-a}$ , and — via a beautiful geometric squeeze called the sandwich theorem — the two limits that unlock every trigonometric derivative, $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\frac{1-\cos x}{x}=0$ . With limits secured, the chapter defines the derivative $f'(x)$ as the limit of the slope of a secant line as it collapses onto a tangent, builds an algebra of derivatives (sum, product, quotient rules), and computes the derivatives of $x^n$ , $\sin x$ , $\cos x$ and $\tan x$ from first principles. By the end you should be able to evaluate any polynomial, rational or trigonometric limit that reduces to a $\tfrac{0}{0}$ form, find a derivative from first principles, and apply the standard results fluently — exactly the toolkit Class 12 calculus, and JEE/NEET, will assume you already own.

Key takeaways

A limit $\lim_{x\to a} f(x) = l$ is the value $f(x)$ is expected to take at $x=a$ , based on the values of $f$ near $a$ — it need not equal $f(a)$ , and $f(a)$ need not even exist.
The left hand limit and right hand limit are the values dictated by points just below $a$ and just above $a$ respectively; the two-sided limit exists only when both agree.
The algebra of limits (Theorem 1) says that if $\lim f(x)$ and $\lim g(x)$ both exist at $x=a$ , then the limit of the sum, difference, product and quotient of $f$ and $g$ equal the sum, difference, product and quotient of the limits (quotient needs the denominator's limit to be non-zero).
The standard result $\lim_{x\to a}\dfrac{x^n-a^n}{x-a} = n a^{n-1}$ (Theorem 2) turns limits involving powers into pure algebra, and holds even for rational $n$ when $a>0$ .
For rational functions of the $\tfrac{0}{0}$ form at $x=a$ , factor out and cancel $(x-a)$ — legitimate because the limiting process never actually sets $x$ equal to $a$ .
Two trigonometric limits, proved by a geometric squeeze (the sandwich theorem): $\lim_{x\to 0}\dfrac{\sin x}{x} = 1$ and $\lim_{x\to 0}\dfrac{1-\cos x}{x} = 0$ .
The derivative of $f$ at $a$ is $f'(a) = \lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}$ — the first principle of differentiation — and geometrically it is the slope of the tangent to $y=f(x)$ at $x=a$ .
The algebra of derivatives gives $(f\pm g)' = f'\pm g'$ , the Leibniz product rule $(uv)' = u'v+uv'$ , and the quotient rule $\left(\dfrac{u}{v}\right)' = \dfrac{u'v-uv'}{v^2}$ .
Standard derivatives: $\dfrac{d}{dx}(x^n) = nx^{n-1}$ , $\dfrac{d}{dx}(\sin x) = \cos x$ , $\dfrac{d}{dx}(\cos x) = -\sin x$ , and $\dfrac{d}{dx}(\tan x) = \sec^2 x$ .
Key idea: a limit is the value a function is heading toward as x approaches a point, and the derivative is exactly the limit of the average rate of change over a shrinking interval — the slope of a secant line collapsing onto the tangent.

The core insight

The derivative $f'(x) = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$ is nothing but the limit of an average rate of change as the interval shrinks to zero, so the entire machinery of limits — left and right hand limits, the algebra of limits, the standard result $\lim\frac{x^n-a^n}{x-a}=na^{n-1}$ , and the sandwich-theorem limits $\lim\frac{\sin x}{x}=1$ and $\lim\frac{1-\cos x}{x}=0$ — feeds directly into an algebra of derivatives (sum, product, quotient rules) that computes the slope of the tangent to any curve.

What this chapter covers

The chapter opens with a physical motivation — the falling stone — that manufactures the idea of an instantaneous rate of change before any formal definition is given. It then formalises the limit: left hand and right hand limits, when a two-sided limit exists, and an algebra of limits (sum, difference, product, quotient) that turns limit-evaluation into ordinary algebra. Two standard results are proved and used relentlessly: $\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$ , which handles every limit built from powers of $x$ , and the trigonometric pair $\lim_{x\to 0}\frac{\sin x}{x}=1$ , $\lim_{x\to 0}\frac{1-\cos x}{x}=0$ , proved by squeezing $\frac{\sin x}{x}$ between $\cos x$ and $1$ (the sandwich theorem). The chapter then pivots to derivatives: $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ , read geometrically as the slope of the tangent line, and generalised to the derivative function $f'(x)$ . An algebra of derivatives mirrors the algebra of limits (sum, product, quotient rules), and the chapter closes by computing, from first principles and then by rule, the derivatives of $x^n$ , general polynomials, and $\sin x$ , $\cos x$ , $\tan x$ .

The core ideas

From average velocity to instantaneous velocity — the intuitive idea

A body dropped from a cliff covers a distance $s = 4.9t^2$ metres in $t$ seconds. The average velocity between two instants $t_1$ and $t_2$ is simply distance travelled divided by time elapsed, a quantity you have computed since middle school.

\bar v = \frac{s(t_2)-s(t_1)}{t_2-t_1}. \qquad (1)

where: $\bar v$ is the average velocity over the interval, and $s(t)$ is the distance function. Computing $\bar v$ for the interval $t_1=1.99$ to $t_2=2$ gives $19.551\ \text{m/s}$ ; for $t_1=1.999$ to $t_2=2$ it gives a number even closer to a fixed value. NCERT's own tables (12.2 and 12.3) push the starting point closer to $t=2$ from both sides — from below (ending at $t=2$ ) and from above (starting at $t=2$ ) — and both sequences of average velocities squeeze toward the same number, somewhere between $19.551$ and $19.649\ \text{m/s}$ . That common value, obtained by letting the time interval shrink to zero, is called the instantaneous velocity at $t=2$ , and it is exactly what the chapter will soon name the derivative of $s$ at $t=2$ . The geometric picture behind this arithmetic is the one that matters most for everything that follows. Plot $s$ against $t$ . The average velocity between $t=2$ and a nearby instant $t=2+h$ is precisely the slope of the secant line joining the two points $(2, s(2))$ and $(2+h, s(2+h))$ on the graph. As $h$ shrinks toward $0$ , the second point slides along the curve toward the first, the secant line rotates, and in the limit it becomes the tangent line to the curve at $t=2$ . The instantaneous velocity is the slope of that tangent — a geometric fact that will reappear, unchanged, as the definition of the derivative in general.

Figure

A distance versus time graph for a falling body showing the secant line through two points on the curve rotating to become the tangent line as the second point slides toward the first, illustrating how average velocity becomes instantaneous velocity

Limits — left hand, right hand, and when they agree

The number a function is expected to take at $x=a$ , based on values of $x$ arbitrarily close to (but not equal to) $a$ , is called the limit of the function at $a$ . Because $x$ can approach $a$ from below or from above, two one-sided notions are needed first.

\lim_{x\to a^-} f(x) \quad \text{and} \quad \lim_{x\to a^+} f(x) \qquad (2)

where: the left hand limit $\lim_{x\to a^-} f(x)$ is the value dictated by points $x$ just less than $a$ , and the right hand limit $\lim_{x\to a^+} f(x)$ is the value dictated by points $x$ just greater than $a$ . The two-sided limit $\lim_{x\to a} f(x)$ exists, and equals that common value, exactly when the left and right hand limits coincide. Consider $f(x) = 1$ for $x \le 0$ and $f(x)=2$ for $x>0$ : the left hand limit at $0$ is $1$ , the right hand limit is $2$ , they disagree, and so $\lim_{x\to 0} f(x)$ simply does not exist — even though $f(0)=1$ is perfectly well defined. This example is the single cleanest illustration in the whole chapter that a function's value at a point and its limit at that point are two entirely different questions.

Figure

A step graph of a function equal to 1 for x less than or equal to zero and equal to 2 for x greater than zero, showing that the left hand limit and right hand limit at x equals zero are different so the two sided limit does not exist

The algebra of limits

Once two limits are known to exist, combinations of the two functions can be handled without recomputing anything from the definition. This is stated (without proof, as NCERT itself does) as a theorem.

\lim_{x\to a}[f(x)\pm g(x)] = \lim_{x\to a} f(x) \pm \lim_{x\to a} g(x), \qquad \lim_{x\to a}[f(x)\,g(x)] = \lim_{x\to a} f(x)\cdot \lim_{x\to a} g(x), \qquad \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)}. \qquad (3)

where: all three identities require $\lim_{x\to a} f(x)$ and $\lim_{x\to a} g(x)$ to individually exist, and the quotient identity additionally needs $\lim_{x\to a} g(x) \ne 0$ . A useful special case: if $g$ is the constant function $\lambda$ , then $\lim_{x\to a}[\lambda f(x)] = \lambda \lim_{x\to a} f(x)$ — constants pull straight out of a limit. This theorem is the reason a limit of a complicated expression can be evaluated piece by piece rather than wrestled with as a whole.

Limits of polynomial and rational functions

Since $\lim_{x\to a} x = a$ trivially (the function $f(x)=x$ approaches $a$ as $x$ approaches $a$ ), repeated use of the product rule in $(3)$ gives $\lim_{x\to a} x^n = a^n$ for every positive integer $n$ , and then term-by-term use of the sum rule gives the limit of any polynomial.

f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n \ \implies\ \lim_{x\to a} f(x) = a_0 + a_1a + a_2a^2 + \cdots + a_na^n = f(a). \qquad (4)

where: the striking conclusion is that for a polynomial, the limit at $a$ is simply the value of the polynomial at $a$ — direct substitution always works. For a rational function $f(x) = \frac{g(x)}{h(x)}$ with $g,h$ polynomials and $h(a)\ne 0$ , the quotient rule in $(3)$ gives $\lim_{x\to a} f(x) = \frac{g(a)}{h(a)}$ just as directly. The interesting case, and the one that dominates exam questions, is $h(a)=0$ . If also $g(a)\ne 0$ , the limit does not exist (the function blows up near $a$ ). If both $g(a)=0$ and $h(a)=0$ — the $\tfrac{0}{0}$ indeterminate form — then $(x-a)$ is a common factor of both $g$ and $h$ ; write $g(x)=(x-a)^k g_1(x)$ and $h(x)=(x-a)^\ell h_1(x)$ with $g_1(a), h_1(a) \ne 0$ , cancel the common power of $(x-a)$ (legitimate since $x\ne a$ throughout the limiting process), and evaluate what remains. If $k>\ell$ the limit is $0$ ; if $k=\ell$ the limit is the finite ratio $g_1(a)/h_1(a)$ ; if $k<\ell$ the limit does not exist.

Limits of trigonometric functions — the sandwich theorem

Some limits cannot be settled by direct substitution or simple factoring; they need a squeeze. If a function $g(x)$ is trapped between two other functions $f(x)\le g(x)\le h(x)$ near $a$ , and $f$ and $h$ both approach the same limit at $a$ , then $g$ is forced to approach that same limit too — this is the sandwich theorem (Theorem 4).

f(x) \le g(x) \le h(x) \ \text{ near } a, \quad \lim_{x\to a} f(x) = l = \lim_{x\to a} h(x) \ \implies\ \lim_{x\to a} g(x) = l. \qquad (5)

where: $f$ , $g$ , $h$ share a common domain around $a$ , and $l$ is the common squeezed value. The chapter applies this once a geometric inequality is proved (in the first derivation below): $\cos x < \frac{\sin x}{x} < 1$ for $0 < x < \frac{\pi}{2}$ . Since $\lim_{x\to 0}\cos x = 1$ and the constant function $1$ trivially has limit $1$ , the sandwich theorem immediately forces $\lim_{x\to 0}\frac{\sin x}{x}=1$ . A companion identity, $1-\cos x = 2\sin^2\!\left(\tfrac{x}{2}\right)$ , converts this into the second standard trigonometric limit, proved in full in the second derivation below.

The derivative — definition and geometric meaning

Generalising the falling-stone calculation to any function $f$ and any point $a$ in its domain gives the formal definition of the derivative.

f'(a) = \lim_{h\to 0} \frac{f(a+h)-f(a)}{h}, \qquad \text{provided the limit exists.} \qquad (6)

where: $f'(a)$ is called the derivative of $f$ at $a$ , and this definition — a limit of a difference quotient — is called differentiation from first principles. The numerator $f(a+h)-f(a)$ is the change in the function's value, and $h$ is the change in $x$ ; the ratio is an average rate of change over the interval from $a$ to $a+h$ , exactly as in $(1)$ , and letting $h\to 0$ converts that average rate into the instantaneous rate at $a$ . Geometrically, if $P=(a, f(a))$ and $Q=(a+h, f(a+h))$ are two points on the graph of $y=f(x)$ , the difference quotient $\frac{f(a+h)-f(a)}{h}$ is exactly the slope of the chord (secant) $PQ$ . As $h\to 0$ , $Q$ slides along the curve toward $P$ and the secant $PQ$ rotates into the tangent line at $P$ ; hence $f'(a)$ equals the slope of the tangent to $y=f(x)$ at the point $(a,f(a))$ — the geometric fact previewed in the falling-stone picture above, now stated for any curve. The same rotating-secant picture drawn earlier for the falling stone applies unchanged to any curve $y=f(x)$ : replace the distance-time graph by the graph of $f$ , replace $A$ by $P=(a,f(a))$ , and replace the moving point by $Q=(a+h, f(a+h))$ , and the secant-to-tangent limit is precisely $f'(a)$ . Letting $a$ vary produces the derivative function $f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ , defined wherever this limit exists (Definition 2). Common notations for the same object are $\frac{d}{dx}f(x)$ , $\frac{dy}{dx}$ (when $y=f(x)$ ), $Df(x)$ , and, for the value at a specific point, $\left.\frac{df}{dx}\right|_{x=a}$ or $\frac{df}{dx}\bigg|_a$ .

The algebra of derivatives

Because the derivative is itself defined through a limit, every algebra-of-limits rule in $(3)$ transfers directly to an algebra of derivatives (Theorem 5), stated here using $u=f(x)$ , $v=g(x)$ for brevity.

(u\pm v)' = u'\pm v', \qquad (uv)' = u'v + uv', \qquad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\ \ (v\ne 0). \qquad (7)

where: the middle identity is the Leibniz product rule: the derivative of a product is not the product of the derivatives, but the first function's derivative times the second, plus the first times the second's derivative. The last identity is the quotient rule: numerator's derivative times denominator, minus numerator times denominator's derivative, all divided by the denominator squared — the order in the numerator matters, since swapping it flips the sign. Both rules let a derivative be computed by combining known building-block derivatives instead of returning to the first-principles limit every time.

Standard derivatives

The single most useful building block is the derivative of $x^n$ , proved in full in the third derivation below.

\frac{d}{dx}(x^n) = nx^{n-1} \ \text{for any positive integer } n; \qquad \frac{d}{dx}(a_nx^n+\cdots+a_1x+a_0) = na_nx^{n-1}+\cdots+2a_2x+a_1. \qquad (8)

where: the second identity (Theorem 7) follows from the first by the sum rule in $(7)$ applied term by term, and it is stated here for a general polynomial of degree $n$ with real coefficients $a_0,\ldots,a_n$ . The result for $x^n$ is stated for any real power in the chapter's remark, though only positive integer $n$ is proved. Alongside this, the fourth derivation below establishes the trigonometric derivatives.

\frac{d}{dx}(\sin x) = \cos x, \qquad \frac{d}{dx}(\cos x) = -\sin x, \qquad \frac{d}{dx}(\tan x) = \sec^2 x. \qquad (9)

where: these three, combined with the algebra of derivatives in $(7)$ , generate the derivative of any combination of trigonometric functions — for instance $\cot x = \frac{\cos x}{\sin x}$ differentiates by the quotient rule to $-\csc^2 x$ , and $\sin^2 x = \sin x\sin x$ differentiates by the product rule to $2\sin x\cos x = \sin 2x$ . The minus sign in $\frac{d}{dx}(\cos x) = -\sin x$ is the single most commonly dropped sign in this entire chapter.

Derivation AThe standard limit for powers of x

The goal of this derivation is to prove Theorem 2 of the chapter, the algebraic result $\lim_{x\to a}\dfrac{x^n-a^n}{x-a} = na^{n-1}$ for any positive integer $n$ , using nothing beyond ordinary polynomial division and the algebra of limits already established. This single result is what makes every polynomial and rational limit involving powers of $x$ tractable without ad-hoc factoring tricks, and it reappears immediately afterward as the engine of the derivative of $x^n$ .

Step 1. Divide the polynomial $x^n - a^n$ by $x - a$ . Polynomial long division confirms the standard factorisation $x^n - a^n = (x-a)\left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + xa^{n-2} + a^{n-1}\right)$ , a sum of exactly $n$ terms, each of total degree $n-1$ in $x$ and $a$ together.

Step 2. Divide both sides by $(x-a)$ , valid since the limiting process only considers $x$ near $a$ but never equal to $a$ . This gives $\dfrac{x^n-a^n}{x-a} = x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + xa^{n-2} + a^{n-1}$ identically for all $x\ne a$ — an exact algebraic identity, not yet a limit statement.

Step 3. Take the limit of both sides as $x\to a$ . The right side is a polynomial in $x$ (for fixed $a$ ), so by the polynomial-limit result in $(4)$ its limit is obtained by direct substitution of $x=a$ : $\lim_{x\to a}\left(x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1}\right) = a^{n-1}+a^{n-2}\cdot a+\cdots+a\cdot a^{n-2}+a^{n-1}$ .

Step 4. Simplify. Every one of the $n$ terms from Step 3 equals $a^{n-1}$ , since $a^{n-1-k}\cdot a^k = a^{n-1}$ for each $k=0,1,\ldots,n-1$ . Adding $n$ copies of $a^{n-1}$ gives $na^{n-1}$ , so $\lim_{x\to a}\dfrac{x^n-a^n}{x-a} = na^{n-1}$ . NCERT remarks that the same formula holds when $n$ is any rational number and $a>0$ , a fact used freely later without re-proof. Verified: for $n=15$ , $a=1$ , all $15$ terms of Step 2 equal $1^{14}=1$ , summing to $15$ , matching $\lim_{x\to1}\frac{x^{15}-1}{x-1}=15$ , used directly in the competitive worked example below.

Go deeper · Derivation B · 1 minThe sandwich theorem proof that sin x over x approaches 1

Derivation BThe sandwich theorem proof that sin x over x approaches 1

The goal here is to prove the two trigonometric limits in $(9)$ 's companion result, $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\frac{1-\cos x}{x}=0$ , starting from a purely geometric comparison of areas in a unit circle and finishing with the sandwich theorem stated in $(5)$ . No series expansion or calculus circularity is used — only elementary geometry of a circular sector and two triangles, exactly as NCERT presents it.

Step 1. Set up the geometric comparison. Let $O$ be the centre of a unit circle ( $OA=1$ ), let $A$ and $C$ be points on the circle with angle $AOC=x$ radians, $0<x<\frac{\pi}{2}$ , and drop perpendiculars: $BA$ perpendicular to $OA$ at $A$ meeting line $OC$ extended at $B$ , and $CD$ perpendicular to $OA$ at $D$ . Triangle $OAC$ sits inside sector $OAC$ , which sits inside triangle $OAB$ , so Area( $\triangle OAC$ ) $<$ Area(sector $OAC$ ) $<$ Area( $\triangle OAB$ ).

Step 2. Compute each area. Triangle $OAC$ has area $\frac12(1)\sin x$ , since $CD=OC\sin x=\sin x$ . The sector has area $\frac{x}{2\pi}\cdot\pi(OA)^2=\frac{x}{2}$ . Triangle $OAB$ has area $\frac12\tan x$ , since $AB=OA\tan x=\tan x$ . Substituting and cancelling $\frac12$ : $\sin x < x < \tan x$ .

Step 3. Convert to the target form. Since $\sin x>0$ on $0<x<\frac{\pi}{2}$ , dividing throughout by $\sin x$ gives $1<\dfrac{x}{\sin x}<\dfrac{1}{\cos x}$ ; taking reciprocals reverses both inequalities to $\cos x < \dfrac{\sin x}{x} < 1$ — inequality $(*)$ used earlier. Since $\sin(-x)=-\sin x$ and $\cos(-x)=\cos x$ , the same inequality holds for $-\frac{\pi}{2}<x<0$ , so it holds on a full punctured neighbourhood of $0$ .

Step 4. Apply the sandwich theorem from $(5)$ : since $\lim_{x\to0}\cos x=1$ and the constant $1$ also has limit $1$ , $\lim_{x\to0}\dfrac{\sin x}{x}=1$ . For the second limit, use $1-\cos x=2\sin^2\!\left(\frac{x}{2}\right)$ : $\dfrac{1-\cos x}{x}=\sin\!\left(\frac{x}{2}\right)\cdot\dfrac{\sin(x/2)}{x/2}$ . As $x\to0$ , $\frac{x}{2}\to0$ , so $\dfrac{\sin(x/2)}{x/2}\to1$ by Step 4's first result, while $\sin(x/2)\to0$ ; by the product rule for limits, $\lim_{x\to0}\dfrac{1-\cos x}{x}=0\times1=0$ . Verified numerically: at $x=0.01$ radians, $\sin x/x = 0.999983\ldots \approx 1$ and $(1-\cos x)/x = 0.005 \approx 0$ , both consistent with the proved limits.

Go deeper · Derivation C · 1 minThe derivative of x to the power n from first principles

Derivation CThe derivative of x to the power n from first principles

The goal of this derivation is to prove Theorem 6, $\dfrac{d}{dx}(x^n)=nx^{n-1}$ for any positive integer $n$ , starting from the first-principles definition $(6)$ and the binomial theorem — the single most-used differentiation formula in the entire syllabus, and the one every polynomial derivative in $(8)$ ultimately rests on. A second, independent proof by induction and the product rule is given at the end, since NCERT presents both and the two illuminate different aspects of the result.

Step 1. Write the first-principles definition for $f(x)=x^n$ : $f'(x) = \lim_{h\to 0}\dfrac{(x+h)^n - x^n}{h}$ .

Step 2. Expand $(x+h)^n$ using the binomial theorem: $(x+h)^n = {}^{n}C_0\,x^n + {}^{n}C_1\,x^{n-1}h + {}^{n}C_2\,x^{n-2}h^2 + \cdots + {}^{n}C_n\,h^n$ . Since $^{n}C_0=1$ , subtracting $x^n$ gives $(x+h)^n - x^n = {}^{n}C_1\,x^{n-1}h + {}^{n}C_2\,x^{n-2}h^2 + \cdots + h^n$ , factored as $h\left(nx^{n-1} + {}^{n}C_2\,x^{n-2}h+\cdots + h^{n-1}\right)$ , using $^{n}C_1=n$ .

Step 3. Substitute into the difference quotient and cancel the common factor of $h$ (valid since $h\ne 0$ throughout the limiting process): $\dfrac{(x+h)^n-x^n}{h} = nx^{n-1} + {}^{n}C_2\,x^{n-2}h + \cdots + h^{n-1}$ . Every term after the first carries at least one positive power of $h$ .

Step 4. Take the limit as $h\to 0$ ; every term with a positive power of $h$ vanishes, leaving $f'(x)=nx^{n-1}$ . As an independent check, induction also works: the base case $n=1$ gives $\frac{d}{dx}(x)=\lim_{h\to0}\frac{(x+h)-x}{h}=1$ ; assuming $\frac{d}{dx}(x^{n-1})=(n-1)x^{n-2}$ and writing $x^n=x\cdot x^{n-1}$ , the product rule from $(7)$ gives $\frac{d}{dx}(x^n) = 1\cdot x^{n-1}+x\cdot(n-1)x^{n-2} = nx^{n-1}$ , matching Step 4 and completing the induction. Verified: for $n=2$ , Step 2 to 4 gives $\frac{(x+h)^2-x^2}{h} = \frac{2xh+h^2}{h} = 2x+h \to 2x$ as $h\to0$ , matching $nx^{n-1}=2x^1$ ; direct product-rule computation of $x\cdot x$ also gives $1\cdot x + x\cdot 1 = 2x$ .

Go deeper · Derivation D · 1 minThe derivative of sin x from first principles

Derivation DThe derivative of sin x from first principles

The goal of this derivation is to compute $\dfrac{d}{dx}(\sin x)$ directly from the definition $(6)$ , using a sum-to-product trigonometric identity and the two standard trigonometric limits proved above, and then to show that an identical method yields $\dfrac{d}{dx}(\cos x)=-\sin x$ — together these two results, combined with the quotient rule, immediately give $\dfrac{d}{dx}(\tan x)=\sec^2x$ as stated in $(9)$ .

Step 1. Write the first-principles definition for $f(x)=\sin x$ : $f'(x) = \lim_{h\to 0}\dfrac{\sin(x+h)-\sin x}{h}$ . Use the sum-to-product identity $\sin A - \sin B = 2\cos\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)$ with $A=x+h$ , $B=x$ : $\sin(x+h)-\sin x = 2\cos\!\left(x+\frac{h}{2}\right)\sin\!\left(\frac{h}{2}\right)$ .

Step 2. Substitute into the difference quotient and regroup the $2$ with the $h$ in the denominator: $f'(x) = \lim_{h\to 0}\cos\!\left(x+\frac{h}{2}\right)\cdot \dfrac{\sin(h/2)}{h/2}$ .

Step 3. Take the limit of each factor separately using the product rule for limits in $(3)$ . As $h\to 0$ , $\cos\!\left(x+\frac{h}{2}\right) \to \cos x$ by direct substitution, and $\frac{h}{2}\to 0$ , so by the sandwich-theorem result proved above, $\dfrac{\sin(h/2)}{h/2} \to 1$ .

Step 4. Multiply the two limits: $f'(x) = \cos x \cdot 1 = \cos x$ , establishing $\dfrac{d}{dx}(\sin x)=\cos x$ . The identical method, starting from $\cos(x+h)-\cos x = -2\sin\!\left(x+\frac{h}{2}\right)\sin\!\left(\frac{h}{2}\right)$ , gives $\dfrac{d}{dx}(\cos x) = -\sin x \cdot 1 = -\sin x$ . Finally, writing $\tan x = \dfrac{\sin x}{\cos x}$ and applying the quotient rule from $(7)$ : $\dfrac{d}{dx}(\tan x) = \dfrac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2 x} = \dfrac{\cos^2x+\sin^2x}{\cos^2x} = \sec^2 x$ , completing the trio in $(9)$ . Verified: at $x=0$ , $\frac{d}{dx}(\sin x)|_{x=0}=\cos 0 = 1$ , matching NCERT's direct computation $\lim_{h\to0}\frac{\sin h}{h}=1$ ; at $x=0$ , $\frac{d}{dx}(\cos x)|_{x=0}=-\sin 0=0$ .

Worked examples

Board level - evaluating a limit that reduces to the 0/0 form (NCERT Example 2)

Evaluate $\displaystyle\lim_{x\to 2} \dfrac{x^3-4x^2+4x}{x^2-4}$ .

Step 1. Check direct substitution first: at $x=2$ , the numerator is $8-16+8=0$ and the denominator is $4-4=0$ , a $\tfrac{0}{0}$ indeterminate form, so direct substitution fails and factoring is needed.

Step 2. Factor both numerator and denominator to expose the common factor $(x-2)$ . The numerator $x^3-4x^2+4x = x(x^2-4x+4) = x(x-2)^2$ . The denominator $x^2-4 = (x+2)(x-2)$ .

Step 3. Cancel the common factor $(x-2)$ , valid because $x\ne 2$ throughout the limiting process: $\dfrac{x(x-2)^2}{(x+2)(x-2)} = \dfrac{x(x-2)}{x+2}$ for $x\ne 2$ . This new expression has no remaining $\tfrac{0}{0}$ obstruction at $x=2$ .

Step 4. Substitute $x=2$ into the simplified expression: $\dfrac{2(2-2)}{2+2} = \dfrac{2(0)}{4} = 0$ . Answer: $\displaystyle\lim_{x\to 2} \dfrac{x^3-4x^2+4x}{x^2-4} = 0$ .

Board level - finding a derivative from first principles (NCERT Example 6)

Find the derivative of $f(x) = 2x^2+3x-5$ at $x=-1$ , and verify that $f'(0)+3f'(-1)=0$ .

Step 1. Write the first-principles definition at $x=-1$ : $f'(-1) = \lim_{h\to0}\dfrac{f(-1+h)-f(-1)}{h}$ . Compute $f(-1+h) = 2(-1+h)^2+3(-1+h)-5 = 2(1-2h+h^2)-3+3h-5 = 2-4h+2h^2-3+3h-5 = 2h^2-h-6$ , and $f(-1) = 2(1)-3-5=-6$ .

Step 2. Subtract and divide: $f'(-1) = \lim_{h\to0}\dfrac{(2h^2-h-6)-(-6)}{h} = \lim_{h\to0}\dfrac{2h^2-h}{h} = \lim_{h\to0}(2h-1) = -1$ , cancelling the common factor of $h$ before taking the limit.

Step 3. Repeat the process at $x=0$ : $f(0+h)=f(h)=2h^2+3h-5$ and $f(0)=-5$ , so $f'(0) = \lim_{h\to0}\dfrac{(2h^2+3h-5)-(-5)}{h} = \lim_{h\to0}\dfrac{2h^2+3h}{h} = \lim_{h\to0}(2h+3) = 3$ .

Step 4. Verify the stated identity: $f'(0)+3f'(-1) = 3+3(-1) = 3-3=0$ , exactly as claimed. Answer: $f'(-1) = -1$ , $f'(0)=3$ , and $f'(0)+3f'(-1)=0$ .

Competitive level (JEE / NEET) - a standard-limit shortcut for a ratio of powers (NCERT Example 3(i))

Evaluate $\displaystyle\lim_{x\to 1}\dfrac{x^{15}-1}{x^{10}-1}$ .

Step 1. Recognise the $\tfrac{0}{0}$ form (both numerator and denominator vanish at $x=1$ ) and rewrite the ratio to expose two instances of the standard limit $\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$ from the first derivation, using $a=1$ : $\dfrac{x^{15}-1}{x^{10}-1} = \dfrac{x^{15}-1}{x-1}\Big/\dfrac{x^{10}-1}{x-1}$ , dividing and multiplying by $(x-1)$ .

Step 2. Split the limit of the quotient using the algebra of limits, since both individual limits exist: $\lim_{x\to1}\dfrac{x^{15}-1}{x^{10}-1} = \dfrac{\lim_{x\to1}\frac{x^{15}-1}{x-1}}{\lim_{x\to1}\frac{x^{10}-1}{x-1}}$ .

Step 3. Apply the standard limit to each piece with $a=1$ : $\lim_{x\to1}\dfrac{x^{15}-1}{x-1} = 15(1)^{14}=15$ and $\lim_{x\to1}\dfrac{x^{10}-1}{x-1} = 10(1)^{9}=10$ .

Step 4. Divide the two results: $\dfrac{15}{10} = \dfrac{3}{2}$ . Answer: $\displaystyle\lim_{x\to 1}\dfrac{x^{15}-1}{x^{10}-1} = \dfrac{3}{2}$ . Timing note: under a minute. The instant a limit of the form $\frac{x^m-1}{x^n-1}$ (or more generally $\frac{x^m-a^m}{x^n-a^n}$ ) appears at $x\to a$ , skip factoring entirely and go straight to the ratio $\frac{m\,a^{m-1}}{n\,a^{n-1}}$ — here $\frac{15(1)^{14}}{10(1)^9}=\frac{15}{10}=\frac32$ in one line. The only trap is applying the standard limit formula when the exponent is not a plain positive integer power of $x$ alone (for example if the base itself were $x^2$ rather than $x$ ) — check the form matches $\frac{x^n-a^n}{x-a}$ exactly before shortcutting.

Competitive level (JEE / NEET) - trigonometric limits by algebraic manipulation (NCERT Example 4)

Evaluate (a) $\displaystyle\lim_{x\to 0}\dfrac{\sin 4x}{\sin 2x}$ and (b) $\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x}$ .

Step 1. For part (a), multiply and divide to create the standard form $\frac{\sin\theta}{\theta}$ in both numerator and denominator: $\dfrac{\sin 4x}{\sin 2x} = \dfrac{\sin 4x}{4x}\cdot\dfrac{2x}{\sin 2x}\cdot\dfrac{4x}{2x} = \dfrac{\sin4x}{4x}\Big/\dfrac{\sin2x}{2x}\cdot 2$ , regrouping so that $4x\to0$ and $2x\to0$ exactly when $x\to0$ .

Step 2. Apply the algebra of limits and the standard result $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ from the second derivation, to each factor separately: $\lim_{x\to0}\dfrac{\sin4x}{4x}=1$ and $\lim_{x\to0}\dfrac{\sin2x}{2x}=1$ , since as $x\to0$ both $4x\to0$ and $2x\to0$ .

Step 3. Combine: $\displaystyle\lim_{x\to0}\dfrac{\sin4x}{\sin2x} = 2\cdot\dfrac{\lim_{x\to0}\frac{\sin4x}{4x}}{\lim_{x\to0}\frac{\sin2x}{2x}} = 2\cdot\dfrac{1}{1} = 2$ . For part (b), write $\tan x = \dfrac{\sin x}{\cos x}$ , so $\dfrac{\tan x}{x} = \dfrac{\sin x}{x}\cdot\dfrac{1}{\cos x}$ .

Step 4. Take the limit of the product using $(3)$ : $\lim_{x\to0}\dfrac{\sin x}{x}=1$ and $\lim_{x\to0}\dfrac{1}{\cos x} = \dfrac{1}{\cos 0}=\dfrac{1}{1}=1$ , so $\lim_{x\to0}\dfrac{\tan x}{x} = 1\times1=1$ . Answer: (a) $\displaystyle\lim_{x\to 0}\dfrac{\sin 4x}{\sin 2x}=2$ and (b) $\displaystyle\lim_{x\to 0}\dfrac{\tan x}{x}=1$ . Timing note: about ninety seconds for both parts together. The reflex to build is: whenever you see $\sin(kx)$ or $\tan(kx)$ over $x$ (or over another $\sin$ or $\tan$ expression), multiply and divide by the coefficient $k$ inside the argument to force the exact pattern $\frac{\sin(kx)}{kx}\to1$ ; never apply $\lim\frac{\sin x}{x}=1$ to $\frac{\sin 4x}{x}$ directly without first inserting the missing factor of $4$ . A frequent slip under time pressure is forgetting the leftover multiplicative constant (the $2$ in part (a)) after the two standard-limit factors have been extracted.

Common traps

Stopping at $\tfrac00$ . A $\tfrac00$ result is not the answer — it signals a common factor $(x-a)$ that must be found and cancelled before substituting again. Never report " $\frac00$ " or "undefined" without first attempting to factor.

Treating $f(a)$ and $\lim_{x\to a} f(x)$ as automatically equal. They coincide for "nice" functions like polynomials, but the piecewise example above shows a function whose value is defined while its limit is not, and vice versa. Always answer the exact question asked.

Ignoring left and right hand limits for piecewise functions. Whenever $f$ changes formula at $x=a$ , compute $\lim_{x\to a^-} f(x)$ and $\lim_{x\to a^+} f(x)$ separately, and only claim the two-sided limit exists if they match exactly.

Forgetting the quotient rule's minus sign and squared denominator. $\left(\frac{u}{v}\right)' = \frac{u'v-uv'}{v^2}$ , not $\frac{u'v+uv'}{v^2}$ and not over plain $v$ — both slips are common under exam pressure.

Misremembering the sign of $\frac{d}{dx}(\cos x)$ . It is $-\sin x$ , not $+\sin x$ ; the derivation above shows the minus sign comes from the identity $\cos A-\cos B=-2\sin(\cdots)\sin(\cdots)$ .

Applying $\lim\frac{x^n-a^n}{x-a}=na^{n-1}$ to a pattern that does not actually match, such as $\frac{x^n-a^m}{x-a}$ with $n\ne m$ , or a base other than plain $x$ . Check exponents and base before shortcutting; when in doubt, factor directly as in the first worked example.

Common trapThe single biggest misconception: a limit is not "just plug in the number"

Students routinely treat $\lim_{x\to a} f(x)$ as a fancy way of writing $f(a)$ , and that habit works often enough on easy problems to go unquestioned — until it silently produces a wrong answer or a $\tfrac00$ dead end. The limit is fundamentally a statement about the values $f(x)$ takes for $x$ near $a$ but never equal to $a$ ; it describes a trend, not an evaluation. This is why cancelling a common factor of $(x-a)$ from numerator and denominator before substituting is legitimate — throughout the entire limiting process $x\ne a$ , so dividing by $(x-a)$ never divides by zero — and it is exactly why a function can have a well-defined limit at a point where it is not even defined (like $h(x)=\frac{x^2-4}{x-2}$ at $x=2$ ), or be defined at a point where the limit does not exist at all (the step-function example above). Whenever a limit problem seems to demand "the value of the function at that point," stop and ask instead: what is $f(x)$ doing as $x$ sneaks up on $a$ from both sides?

You should be able to…

Evaluate $\lim_{x\to a} f(x)$ for polynomial functions by direct substitution, using the result that a polynomial's limit equals its value at $a$ .
Recognise a $\tfrac{0}{0}$ indeterminate form in a rational function's limit, factor out the common $(x-a)$ , and evaluate the simplified expression.
State the left hand limit and right hand limit of a piecewise function, and correctly decide whether the two-sided limit exists.
Apply the algebra of limits (sum, difference, product, quotient) to break a compound limit into simpler pieces that can be evaluated separately.
Use the standard result $\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}$ to shortcut limits built from powers, including ratios like $\frac{x^m-1}{x^n-1}$ .
Evaluate trigonometric limits at $0$ using $\lim_{x\to0}\frac{\sin x}{x}=1$ and $\lim_{x\to0}\frac{1-\cos x}{x}=0$ , inserting missing multiplicative constants where needed.
Compute a derivative from first principles, $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ , for polynomial and simple trigonometric functions.
Interpret $f'(a)$ geometrically as the slope of the tangent line to $y=f(x)$ at $x=a$ , and connect it to the idea of instantaneous rate of change.
Apply the sum, Leibniz product, and quotient rules to differentiate combinations of functions without returning to first principles every time.
Differentiate $x^n$ , general polynomials, $\sin x$ , $\cos x$ and $\tan x$ fluently, both from the standard formulas and, when asked, from first principles.

Ask the tutor: Want to work through a first-principles derivative or a tricky 0/0 limit together, step by step?

Why does the limit at x equals a not care what happens exactly at x equals a?

Because a limit describes a trend in the values of f(x) for x close to a but never equal to a, not a single evaluation at a. The formal left and right hand limits are both built entirely from points near a, deliberately excluding a itself. This is exactly why a function can have a limit at a point where it is not even defined, such as $\frac{x^2-4}{x-2}$ at $x=2$ , and why a function can be defined at a point while its limit there fails to exist, as in the step-function example in the core ideas above. The value f(a), if it exists, is a separate fact that the limit simply does not consult.

Why can we cancel the factor x minus a from the numerator and denominator when a limit gives 0 over 0?

Because throughout the entire limiting process x is only ever taken near a, never equal to a, so $x-a$ is never actually zero and dividing by it is always legitimate. Once you write $g(x)=(x-a)g_1(x)$ and $h(x)=(x-a)h_1(x)$ , the identity $\frac{g(x)}{h(x)}=\frac{g_1(x)}{h_1(x)}$ holds for every $x\ne a$ , which is exactly the set of points a limit examines. After cancelling, you evaluate the simplified expression at $x=a$ using ordinary substitution, since $g_1$ and $h_1$ are typically well behaved there even though the original ratio was not.

Is dy by dx a fraction, and can I split it into dy and dx separately?

Not at this stage. $\frac{dy}{dx}$ is a single symbol denoting the derivative, defined as the limit $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ , and it should be treated as one indivisible object here, not as an actual fraction with a numerator dy and a denominator dx that can be manipulated separately. It behaves suggestively like a fraction (the quotient and product rules even look like fraction algebra), which is a useful mnemonic, but formally justifying dy and dx as separate quantities that can be split apart belongs to later, more careful treatments of calculus, not to this first-principles chapter.

Why is the limit of sin x over x equal to 1 only when x is measured in radians?

The geometric proof above rests on the inequality $\sin x < x < \tan x$ , which compares an actual arc length (the radian measure $x$ ) against straight-line lengths $\sin x$ and $\tan x$ on a unit circle — the comparison is only valid because radian measure is itself defined as arc length divided by radius. If $x$ were measured in degrees, the same geometric picture would introduce an extra constant factor of $\frac{\pi}{180}$ , so the limit would instead approach $\frac{\pi}{180}$ , not $1$ . Every trigonometric limit and derivative in this chapter silently assumes radian measure for exactly this reason.

What is the actual difference between average velocity and instantaneous velocity, and how does the derivative connect them?

Average velocity is distance travelled divided by time elapsed over some finite interval, such as from $t=1.99$ to $t=2$ seconds — it is a single number attached to an interval, not to an instant. Instantaneous velocity is what that average velocity settles down to as the interval is shrunk toward a single instant, and the derivative $s'(t)=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}$ is the formal name for exactly this limiting process. In other words, instantaneous velocity is not a different kind of quantity from average velocity; it is the limit of average velocities taken over intervals that shrink to zero length, which is precisely why the chapter introduces limits before it ever defines the derivative.