Limits and Derivatives Class 11 Notes: Formulas and Tips

These limits and derivatives class 11 notes compress NCERT Chapter 12 into definitions students actually understand, every formula you need, fully worked examples with original numbers, and examiner-minded tips for last-minute revision. They are part of our broader Class 11 Mathematics notes, which sit inside the full Class 11 notes collection on this site.

Work top to bottom: first the derivative’s meaning (a rate of change), then the limit machinery it rests on, then the algebra rules for both. Each section carries the NCERT page number so you can jump back to the textbook where a point needs more depth.

The official NCERT PDF for Chapter 12 (kemh112.pdf) is free to download from ncert.nic.in if you want the full chapter alongside these notes.

The Intuitive Idea: Velocity as a Derivative

The chapter opens with a falling body. Drop a stone from a tall cliff and the distance it covers in \( t \) seconds is \( s = 4.9t^2 \) metres (NCERT, p. 218). The question: what is the velocity at the exact instant \( t = 2 \)?

Average velocity between two instants \( t_1 \) and \( t_2 \) is the distance travelled divided by the time interval: \( v = \frac{s(t_2)-s(t_1)}{t_2-t_1} \). For the first two seconds that gives \( \frac{19.6-0}{2-0} = 9.8 \) m/s (NCERT, p. 218).

The trick is to shrink the interval towards \( t = 2 \). Intervals ending at \( t = 2 \) give 14.7, 19.11, 19.355, 19.551 m/s as the start point creeps up to 1.99 s. Intervals starting at \( t = 2 \) give 20.09, 19.845, 19.649 m/s as the end point drops towards 2.01 s (NCERT, pp. 219–220).

Both sequences press towards one number near 19.6 m/s. That shared target is the instantaneous velocity at \( t = 2 \), and it is what we call the derivative of \( s = 4.9t^2 \) at \( t = 2 \) (NCERT, p. 220).

The image (Fig 12.1) translates the numbers into geometry on the \( s \)-\( t \) curve. Each ratio \( \frac{CB}{AC} \) is the slope of a chord; as the time interval collapses, the chords slide into the tangent at the point, so the instantaneous velocity equals the tangent’s slope (NCERT, p. 220).

Quick check with original numbers. A car’s position is \( s(t) = 6t^2 – 4t \) metres after \( t \) seconds. Average velocity from \( t=2 \) to \( t=2.1 \): \( \frac{s(2.1)-s(2)}{0.1} = \frac{18.06-16}{0.1} = 20.6 \) m/s. From \( t=2 \) to \( t=2.01 \): \( \frac{16.2006-16}{0.01} = 20.06 \) m/s. The values close in on 20 m/s, so the instantaneous velocity at \( t=2 \) is 20 m/s.

Real-life application — estimating speed from distance data. The same “known values, find the rate” pattern shows up when reservoir managers predict an overflow from water-depth readings, when rocket engineers compute the precise launch velocity of a satellite from its height over time, and when financial analysts track a stock’s swing from its present value (NCERT, p. 240).

Analysing how one quantity changes with another is also the core idea behind the statistics chapter notes.

Limits: Formal Definition and One-Sided Limits

A limit asks what value \( f(x) \) should take at \( x = a \), based purely on the values the function takes near \( a \). For \( h(x) = \frac{x^2-4}{x-2} \), \( x \neq 2 \), the function is undefined at 2, yet plugging in 1.99, 1.999, 2.001, 2.01 always lands near 4. We write \( \lim_{x \to 2} h(x) = 4 \) (NCERT, pp. 221–222).

The graph above makes it visual: the line walks right up to a gap at \( x = 2 \), and the missing value would be 4. The limit is the value the function heads towards, whether or not the function actually reaches it there.

There are two directions from which \( x \) can approach \( a \): from the left (values smaller than \( a \)) or from the right (values larger than \( a \)). These give the left-hand limit \( \lim_{x \to a^-} f(x) \) and the right-hand limit \( \lim_{x \to a^+} f(x) \) (NCERT, p. 222).

When one-sided limits disagree. Take \( f(x) = 1 \) for \( x \le 0 \) and \( f(x) = 2 \) for \( x \gt 0 \). The left-hand limit at 0 is 1, the right-hand limit is 2; because they differ, \( \lim_{x \to 0} f(x) \) does not exist — even though \( f(0) = 1 \) is perfectly defined (NCERT, p. 222).

The jump graph above is Illustration 9 (NCERT, p. 227): \( f(x) = x-2 \) for \( x \lt 0 \), \( f(0) = 0 \), \( f(x) = x+2 \) for \( x \gt 0 \). Values from the left approach −2, values from the right approach +2, and no single limit exists.

Illustration 10 (NCERT, p. 228) makes a second point: the value of a function and its limit at a point can be different. There \( f(1) = 0 \) but \( \lim_{x \to 1} f(x) = 3 \), because nearby behaviour — dictated by the branch \( x + 2 \) — is what a limit records.

Key Definitions: Limit, One-Sided Limits, Derivative

Quick-reference table for the central terms (definitions appear in NCERT, pp. 222 and 240).

Algebra of Limits: Rules for Sum, Product, Quotient

Theorem 1 (NCERT, p. 228): when both \( \lim_{x \to a} f(x) \) and \( \lim_{x \to a} g(x) \) exist, limits respect the four arithmetic operations.

• Sum: \( \lim_{x \to a}[f(x)+g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \)
• Difference: \( \lim_{x \to a}[f(x)-g(x)] = \lim_{x \to a} f(x) – \lim_{x \to a} g(x) \)
• Product: \( \lim_{x \to a}[f(x)\cdot g(x)] = \left(\lim_{x \to a} f(x)\right)\left(\lim_{x \to a} g(x)\right) \)
• Quotient: \( \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \), only when \( \lim_{x \to a} g(x) \neq 0 \)

Special case of the product rule: a constant factor pulls out, \( \lim_{x \to a}[\lambda f(x)] = \lambda \lim_{x \to a} f(x) \) (NCERT, p. 228).

Why the quotient needs a non-zero denominator limit. A quotient’s behaviour is only predictable when the denominator settles away from 0. If \( \lim g(x) = 0 \), the fraction can blow up, oscillate, or approach a finite number depending on how fast numerator and denominator each vanish — one rule cannot decide all those cases.

Limits of Polynomial and Rational Functions

Because \( \lim_{x \to a} x = a \), multiplying step by step gives \( \lim_{x \to a} x^n = a^n \). Adding the terms of a polynomial then yields the headline result: the limit of a polynomial at \( a \) is simply \( f(a) \) (NCERT, p. 229).

Theorem 2 (NCERT, p. 232): for a positive integer \( n \), \( \lim_{x \to a}\frac{x^n – a^n}{x-a} = na^{n-1} \). The remark extends the result to rational \( n \) when \( a \) is positive — this is what later lets you differentiate expressions like \( \sqrt{x} \) by hand.

For a rational function \( \frac{g(x)}{h(x)} \), first try substituting \( x = a \). If \( h(a) \neq 0 \), the limit is simply \( \frac{g(a)}{h(a)} \). Trouble starts when \( h(a) = 0 \):

• If \( g(a) \neq 0 \) — numerator stays non-zero while the denominator vanishes — the limit does not exist.
• If \( g(a) = 0 \) too — the \( \frac{0}{0} \) form — factor both polynomials, cancel the shared factor, then substitute (NCERT, pp. 229–231).

Example where the limit does not exist: \( \lim_{x \to 1}\frac{x+3}{(x-1)(x+2)} \). Substituting gives \( \frac{4}{0} \) — the numerator settles at 4 while the denominator collapses to 0, so the quotient has no finite limit.

Misconception autopsy — why substitution fails for \( \frac{0}{0} \). The form 0/0 is not a number, so the calculator-style answer “0/0 = 0” is meaningless. Two ratios that both look like 0/0 can genuinely approach different limits: \( \frac{x^2-4}{x-2} \to 4 \) while \( \frac{x-2}{x^2-4} \) has no finite limit at 2. Substitution cannot decide — you must identify the factor causing the vanishing and cancel it first.

Standard Trigonometric Limits and the Sandwich Theorem

Two structural theorems come first. Theorem 3: if \( f(x) \le g(x) \) and both limits exist, then \( \lim f(x) \le \lim g(x) \), so limits preserve order (NCERT, p. 234).

The Sandwich Theorem (Theorem 4) builds on this: if \( f(x) \le g(x) \le h(x) \) and \( \lim f(x) = \lim h(x) = l \), then \( \lim g(x) = l \) (NCERT, p. 234).

The two graphs above show the intuition: a function trapped between two others that meet at a common limit is forced to that same limit.

The key inequality (NCERT, p. 235): \( \cos x \lt \frac{\sin x}{x} \lt 1 \) for \( 0 \lt |x| \lt \frac{\pi}{2} \). Its geometric proof compares areas in the unit circle: triangle \( OAC \) is smaller than sector \( OAC \), which is smaller than triangle \( OAB \). That yields \( OA\sin x \lt OA\cdot x \lt OA\tan x \), i.e.

\( \sin x \lt x \lt \tan x \); dividing by \( \sin x \) (positive here) and taking reciprocals gives the inequality.

Since \( \cos x \to 1 \) as \( x \to 0 \), the Sandwich Theorem squeezes \( \frac{\sin x}{x} \) down to 1 (NCERT, p. 235). The companion limit follows from \( 1 – \cos x = 2\sin^2\left(\frac{x}{2}\right) \): \( \frac{1-\cos x}{x} = \frac{\sin(x/2)}{x/2}\cdot\sin(x/2) \to 1\cdot 0 = 0 \) (NCERT, p. 236).

\[ \lim_{x \to 0}\frac{\sin x}{x} = 1,\qquad \lim_{x \to 0}\frac{1-\cos x}{x} = 0 \]

Memory device: “Match the sine to the floor.” For \( \frac{\sin x}{x} \to 1 \), the argument inside sine and the denominator must be the same quantity heading to 0. That is why \( \frac{\sin(7x)}{7x} \to 1 \) but \( \frac{\sin x}{x^2} \) does not.

Whenever you rescale, write the argument and the denominator identically (both \( 5x \), both \( 3x \)) before applying the standard limit.

The trigonometric limit work in this section draws on the identities covered in the trigonometric functions notes.

Derivatives: Definition and First Principle

Definition 1 (NCERT, p. 240): the derivative of \( f \) at a point \( a \) is \( f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h} \), provided this limit exists. It quantifies how \( f \) changes at \( a \) as \( x \) changes.

Definition 2 / first principle (NCERT, p. 242): the derivative function \( f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \), defined wherever the limit exists. Common notations are \( f'(x) \), \( \frac{dy}{dx} \), and \( Df(x) \) — all name the same rate of change.

The image carries the geometric meaning: \( \frac{f(a+h)-f(a)}{h} \) is the slope of the chord \( PQ \). As \( h \to 0 \), Q slides into P and the chord becomes the tangent, so the derivative is the slope of the tangent at that point — \( f'(a) = \tan\psi \) (NCERT, p. 242).

This tangent-slope idea connects to the coordinate geometry you meet in the three-dimensional geometry notes.

Algebra of Derivatives: Product and Quotient Rules

Theorem 5 (NCERT, p. 244): when derivatives exist in a common domain, the following rules hold.

• Sum: \( \frac{d}{dx}[f+g] = \frac{df}{dx} + \frac{dg}{dx} \)
• Difference: \( \frac{d}{dx}[f-g] = \frac{df}{dx} – \frac{dg}{dx} \)
• Product rule: \( \frac{d}{dx}[fg] = \frac{df}{dx}\cdot g + f\cdot\frac{dg}{dx} \)
• Quotient rule: \( \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{\frac{df}{dx}\cdot g – f\cdot\frac{dg}{dx}}{g^2} \), needing \( g \neq 0 \)

With \( u = f(x) \) and \( v = g(x) \), these become the Leibnitz rules (NCERT, p. 245): \( (uv)’ = u’v + uv’ \) and \( \left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2} \).

Why the product rule is not “differentiate each factor”. When both factors vary, the total change of the product gets a contribution from \( u \) changing while \( v \) stays, plus a contribution from \( v \) changing while \( u \) stays. Those two strips of growth are \( u’v + uv’ \).

The wrong guess \( u’v’ \) only counts the tiny corner where both change at once, missing the two main contributions.

Derivatives of Standard Functions

Each derivative below is proved from the first principle together with the standard trigonometric limits (NCERT, pp. 245–247).

For example, \( \frac{d}{dx}\sin x = \lim_{h \to 0}\frac{\sin(x+h)-\sin x}{h} = \cos x \cdot 1 \), using \( \sin A – \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \) and \( \frac{\sin(h/2)}{h/2} \to 1 \) (NCERT, p. 247).

Limits and Derivatives Class 11 Notes: Master Formula Sheet

One compact grid of everything from the chapter (formulas appear across NCERT, pp. 228, 232, 235–236, 244–245, 247).

Worked Examples: Limits and Derivatives

Common Mistakes to Avoid

Five chapter-specific errors cost marks every year. The table pairs each error with the correct rule and a quick self-check.

Spotlight on the cancellation error. Students write \( \lim_{x \to 2}\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{(x-2)} = x+2 \) and then treat \( x+2 \) as if it were the original function everywhere. The correction: \( \frac{x^2-4}{x-2} \) is not defined at \( x = 2 \). Only after cancelling — valid precisely because \( x \neq 2 \) in the limiting process — do we reach \( x+2 \), whose limit is \( 4 \).

So the limit equals 4, but \( f(2) \) does not even exist.

Exam Notes: What CBSE Looks For

• Rational-function limits: substitute first, then factorise for the \( \frac{0}{0} \) cases (pattern of Exercise 12.1, Q2–Q8).
• Trigonometric limits: the two standard limits, with angles in radians (Exercise 12.1, Q13–Q21).
• One-sided limits and piecewise functions decide whether a limit exists (Exercise 12.1, Q23–Q30).
• First-principle derivatives for simple polynomials and trigonometric functions (Exercise 12.2, Q4 and Q10; Miscellaneous Q1, Q14).
• Method marks matter: writing the correct setup — the quotient rule formula, the standard-limit shape — earns marks even if the final simplification slips.
• Write “as \( x \neq a \)” next to each cancellation inside a limit; it shows you understand the limiting condition.

For quick reference back to the full curriculum, the CBSE notes index lists every chapter page on this site.

Revision Summary: One-Page Recap

A recall aid: concept, the rule in one line, and where the textbook proves it.

Frequently Asked Questions

Why is the limit of \( \frac{\sin x}{x} \) equal to 1, and how do I use it?

The Sandwich Theorem forces it. For \( 0 \lt |x| \lt \frac{\pi}{2} \), the unit-circle area comparison gives \( \cos x \lt \frac{\sin x}{x} \lt 1 \). Since \( \cos x \to 1 \) as \( x \to 0 \), the squeezed function must also approach 1. To use it, the angle inside sine must equal the denominator and both must tend to 0 (all in radians), e.g.

\( \frac{\sin 7x}{7x} \to 1 \).

How do I find a limit when direct substitution gives 0/0?

Do not stop — factor the numerator and denominator, cancel the common factor (valid because \( x \neq a \) in the limit), and substitute again. For trigonometric forms, rewrite with the standard limits, like turning \( \frac{1-\cos x}{x} \) into \( \frac{\sin(x/2)}{x/2}\cdot\sin(x/2) \).

What is the difference between the derivative at a point and the derivative function?

The derivative at a point \( f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h} \) is a single number: the rate of change at that one spot.

The derivative function \( f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \) is defined wherever that limit exists and assigns a slope to every point in its domain at once.

When should I use the product rule versus the quotient rule?

Use the product rule when the function is one expression times another, \( uv \) — for example \( (3x+2)(x^2-1) \). Use the quotient rule when one expression is divided by another, \( \frac{u}{v} \), requiring \( v \neq 0 \). The quotient rule is the product rule’s close cousin: \( \frac{u’v – uv’}{v^2} \), with a minus sign and the square of the denominator.

How do I apply the first principle to find the derivative of a function?

Start from \( f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \). Substitute \( x+h \) into the function, subtract \( f(x) \), simplify and divide through by \( h \), then let \( h \to 0 \). With \( f(x) = 2x^3 – 3x \), expanding \( (x+h)^3 \) and cancelling \( h \) leaves \( 6x^2 + 6xh + 2h^2 – 3 \), whose limit is \( 6x^2 – 3 \).

Reference: NCERT Class 11 Mathematics textbook, chapter Limits and Derivatives.