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Limits and Derivatives Class 11 Formulas

Need the Limits and Derivatives Class 11 formulas in one place? This sheet covers the limits side — left and right hand limits, the algebra of limits, polynomial and rational-function limits, and the standard trigonometric limits — and the derivatives side — the first-principle definition, the sum, product and quotient rules, and the derivatives of standard functions.

Every entry gives the meaning of each symbol, the condition under which the formula holds, and a one-line note on when to use it. Original worked examples and the chapter’s common mistakes follow. For the reasoning behind the rules, work through the NCERT chapter; for the wider set, keep the Class 11 maths formulas hub beside you.

Formulas at a Glance

Purpose Formula
Limit of a function (exists when the one-sided limits are equal) \( \lim_{x \to a} f(x) = l \)
Sum or difference of limits \( \lim_{x \to a}[f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \)
Product of limits \( \lim_{x \to a}[f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \)
Quotient of limits (denominator limit non-zero) \( \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \)
Constant multiple of a limit \( \lim_{x \to a}[\lambda f(x)] = \lambda \lim_{x \to a} f(x) \)
Limit of a power of x \( \lim_{x \to a} x^n = a^n \)
Limit of a polynomial (substitute the point) \( \lim_{x \to a} f(x) = f(a) \)
Limit of a rational function (denominator non-zero at the point) \( \lim_{x \to a}\frac{g(x)}{h(x)} = \frac{g(a)}{h(a)} \)
Standard limit for a difference of like powers \( \lim_{x \to a}\frac{x^n – a^n}{x – a} = n a^{n-1} \)
Standard limit: sine over its angle (x in radians) \( \lim_{x \to 0}\frac{\sin x}{x} = 1 \)
Standard limit: 1 minus cos x over x \( \lim_{x \to 0}\frac{1 – \cos x}{x} = 0 \)
Derived: tangent over x (from the sine limit) \( \lim_{x \to 0}\frac{\tan x}{x} = 1 \)
Sandwich theorem: equal outer limits force the middle limit \( f(x) \le g(x) \le h(x),\; \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = l \Rightarrow \lim_{x \to a} g(x) = l \)
Derivative of f at a point a \( f'(a) = \lim_{h \to 0}\frac{f(a+h) – f(a)}{h} \)
Derivative function from first principles \( f'(x) = \lim_{h \to 0}\frac{f(x+h) – f(x)}{h} \)
Slope of the tangent at a point \( f'(a) = \tan \psi \)
Sum or difference rule \( (u \pm v)’ = u’ \pm v’ \)
Product rule (Leibnitz rule) \( (uv)’ = u’v + uv’ \)
Quotient rule (denominator non-zero) \( \left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2} \)
Power rule \( \frac{d}{dx}(x^n) = n x^{n-1} \)
Derivative of a constant \( \frac{d}{dx}(c) = 0 \)
Derivative of sine \( \frac{d}{dx}(\sin x) = \cos x \)
Derivative of cosine \( \frac{d}{dx}(\cos x) = -\sin x \)
Derivative of tangent \( \frac{d}{dx}(\tan x) = \sec^2 x \)
Derivative of cotangent \( \frac{d}{dx}(\cot x) = -\cosec^2 x \)
Derivative of 1 over x (from first principles) \( \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \)
Derivative of a polynomial \( \frac{d}{dx}(a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0) = n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 \)

All Formulas, Grouped by Topic

Grouped under the same sub-topics as the chapter. Results that are derived in worked examples rather than stated as boxed theorems are labelled ‘derived’. Formulas follow the Rationalised NCERT Class 11 Mathematics textbook; you can verify any expression against the chapter PDF on the official NCERT site.

Limits

As x moves towards a from the left, the values of f may push towards one number; from the right, possibly another. The left hand limit and right hand limit record these two one-sided behaviours:

\[ \lim_{x \to a^-} f(x) \quad \text{(left hand limit)} \qquad \lim_{x \to a^+} f(x) \quad \text{(right hand limit)} \]

The limit of f at a exists exactly when the two one-sided limits exist and are equal:

\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = l \quad \Rightarrow \quad \lim_{x \to a} f(x) = l \]

Notice that the limit does not need the function to be defined at a. The graph of \( y = \frac{x^2-4}{x-2} \) below shows a function with a missing point at x = 2 — the limit is still 4.

Graph of a rational function shaped like a straight line with a missing point at x = 2, where the limit equals 4
Fig 12.2: Graph of \( y = \frac{x^2-4}{x-2} \) — the point at x = 2 is missing, but the limit is 4. Source: NCERT

Two defining situations to recognise: when the one-sided limits differ, the limit does not exist (Fig 12.6); and a limit can exist even when it disagrees with the function value (Fig 12.7).

Graph of a piecewise function that jumps from level minus 2 to level 2 at x = 0, so the two one-sided limits differ
Fig 12.6: Left hand limit \( -2 \) and right hand limit \( 2 \) at x = 0 do not coincide, so the limit at 0 does not exist. Source: NCERT
Graph of a piecewise function with an isolated point at height 0 at x = 1 while the curve approaches height 3
Fig 12.7: The value of the function at x = 1 is 0, while the limit as x tends to 1 is 3. Source: NCERT

Algebra of Limits

When both limits exist, the limiting process respects the four basic operations (NCERT, p. 228):

\[ \lim_{x \to a}[f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \]

\[ \lim_{x \to a}[f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \]

\[ \lim_{x \to a}\frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, \quad \lim_{x \to a} g(x) \neq 0 \]

A constant factor is a special case of the product law:

\[ \lim_{x \to a}[\lambda f(x)] = \lambda \lim_{x \to a} f(x) \]

Limits of Polynomials and Rational Functions

For a polynomial, substitution is enough — the limit equals the value of the polynomial at that point:

\[ \lim_{x \to a} x^n = a^n, \qquad \lim_{x \to a} f(x) = f(a) \]

For a rational function \( f(x) = \frac{g(x)}{h(x)} \) with \( h(x) \neq 0 \), substitute first. If the denominator is non-zero at a, the limit is the quotient of the values:

\[ \lim_{x \to a}\frac{g(x)}{h(x)} = \frac{g(a)}{h(a)}, \quad h(a) \neq 0 \]

A 0/0 result means ‘not yet evaluated’: factor the common factor (x minus a) out of both polynomials, cancel, and substitute again. If the denominator vanishes while the numerator does not, the limit does not exist.

The standard limit below handles the equal-powers pattern in one step (NCERT, p. 233):

\[ \lim_{x \to a}\frac{x^n – a^n}{x – a} = n a^{n-1}, \quad n \text{ a positive integer; also any rational } n \text{ with } a \gt 0 \]

Limits of Trigonometric Functions

The two standard trigonometric limits, with x measured in radians (NCERT, p. 235):

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

The first is proved by sandwiching \( \frac{\sin x}{x} \) between \( \cos x \) and 1:

\[ \cos x \lt \frac{\sin x}{x} \lt 1 \quad \text{for } 0 \lt |x| \lt \frac{\pi}{2} \]

The chain \( \sin x \lt x \lt \tan x \) behind that inequality comes from comparing the areas of the triangle, sector and larger triangle in Fig 12.10; the second limit uses \( 1 – \cos x = 2\sin^2\frac{x}{2} \).

Unit circle with an inscribed triangle, a circular sector and a larger right triangle whose areas are compared to prove an important trig inequality
Fig 12.10: Area comparison in the unit circle giving \( \sin x \lt x \lt \tan x \) for \( 0 \lt x \lt \frac{\pi}{2} \). Source: NCERT

Sandwich theorem: if g is trapped between two functions whose limits agree, its limit is forced to the same value:

\[ f(x) \le g(x) \le h(x) \quad \text{and} \quad \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = l \quad \Rightarrow \quad \lim_{x \to a} g(x) = l \]

Derived from \( \frac{\sin x}{x} = 1 \) by writing \( \tan x = \frac{\sin x}{\cos x} \):

\[ \lim_{x \to 0}\frac{\tan x}{x} = 1 \]

Derivatives

The distance-time curve in Fig 12.1 is the motivating picture: average velocities over shrinking time intervals approach the slope of the tangent — that limiting slope is the derivative.

Distance-time graph of a falling body whose chords of shrinking width approach the tangent line at the point of interest
Fig 12.1: Distance-time graph of a falling body — the average-velocity ratios approach the slope of the tangent at A. Source: NCERT

The derivative of f at a point a is the rate of change of f at a, defined as the limit (NCERT, p. 240):

\[ f'(a) = \lim_{h \to 0}\frac{f(a+h) – f(a)}{h} \]

The same limit written at a general point x is the first-principle definition (NCERT, p. 242):

\[ f'(x) = \lim_{h \to 0}\frac{f(x+h) – f(x)}{h} \]

Usual notations: \( f'(x) \), \( \frac{df}{dx} \), \( \frac{dy}{dx} \) when \( y = f(x) \), and \( D(f(x)) \). The geometric meaning: f'(a) is the slope of the tangent at P, so \( f'(a) = \tan \psi \) (NCERT, p. 242).

Curve with a chord through two close points P and Q rotating toward the tangent at P as the second point approaches
Fig 12.11: As Q moves towards P, the chord PQ tends to the tangent at P and the difference quotient tends to \( f'(a) = \tan \psi \). Source: NCERT

Algebra of Derivatives

Let \( u = f(x) \) and \( v = g(x) \). The sum, product and quotient rules (NCERT, p. 244):

\[ (u \pm v)’ = u’ \pm v’ \]

\[ (uv)’ = u’v + uv’ \quad \text{(product rule / Leibnitz rule)} \]

\[ \left( \frac{u}{v} \right)’ = \frac{u’v – uv’}{v^2} \quad \text{(quotient rule, } v \neq 0 \text{ at the point)} \]

Derivatives of Standard Functions

Power rule and the polynomial rule (NCERT, p. 245):

\[ \frac{d}{dx}(x^n) = n x^{n-1} \]

\[ \frac{d}{dx}\left( a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \right) = n a_n x^{n-1} + (n-1)a_{n-1}x^{n-2} + \cdots + a_1 \]

The standard derivatives collected in the chapter summary (NCERT, p. 255), with the tangent and cotangent results from the worked examples:

\[ \frac{d}{dx}(c) = 0 \quad (c \text{ constant; Example 11}), \qquad \frac{d}{dx}(\sin x) = \cos x, \qquad \frac{d}{dx}(\cos x) = -\sin x \]

\[ \frac{d}{dx}(\tan x) = \sec^2 x \quad \text{(Example 17)}, \qquad \frac{d}{dx}(\cot x) = -\cosec^2 x \quad \text{(Example 21)} \]

\[ \frac{d}{dx}\left( \frac{1}{x} \right) = -\frac{1}{x^2} \quad \text{(Example 12)} \]

What Each Symbol Means

Symbol What it means Unit / nature
\( x \) The independent variable of the function Real number
\( a \) The point that x approaches (in limits) or at which the derivative is evaluated Real number
\( h \) Small increment in x in the definition of the derivative; h approaches 0 with h not equal to 0 Real number
\( f, g, h, u, v \) Real-valued functions of x Real numbers
\( l \) The limiting value that f(x) approaches Real number
\( \lambda \) A fixed real constant multiplying a function Real number
\( n \) Exponent in the power limit and the power rule Positive integer (any real power in the power rule)
\( a_0, a_1, \ldots, a_n \) Coefficients of a polynomial Real numbers
\( f'(x), \frac{dy}{dx}, \frac{d}{dx}(f(x)) \) Derivative of f with respect to x — the rate of change of f at x Rate of change (a real number)
\( \psi \) Angle the tangent to the curve y = f(x) makes with the x-axis Radians
\( \sin x, \cos x, \tan x, \cot x, \sec x, \cosec x \) Trigonometric functions of the angle x Dimensionless ratios; x in radians

When to Use Each Formula

Formula or rule Use it when
\( \lim_{x \to a} f(x) \) with one-sided limits A question asks whether a limit exists, especially for a piecewise function. The limit exists only when both one-sided limits exist and are equal.
Algebra of limits You want to split a sum, product or quotient into simpler limits. Both limits must exist; the quotient law also needs the denominator limit non-zero.
\( \lim_{x \to a} f(x) = f(a) \) f is a polynomial. Just substitute the point a — no factoring.
\( \lim_{x \to a}\frac{g(x)}{h(x)} = \frac{g(a)}{h(a)} \) f is a rational function and the denominator is non-zero at a. Substitute directly.
Factor and cancel Substitution gives 0/0. Factor (x minus a) out of numerator and denominator, cancel, then substitute again.
\( \lim_{x \to a}\frac{x^n – a^n}{x – a} = n a^{n-1} \) You meet exactly this pattern of equal powers over the difference of bases, or a substitution can create it. n a positive integer; also rational n with a positive.
\( \lim_{x \to 0}\frac{\sin x}{x} = 1 \) A trigonometric limit of the 0/0 type. x must be in radians; rescale so the sine’s angle matches the denominator.
\( \lim_{x \to 0}\frac{1 – \cos x}{x} = 0 \) A 0/0 limit containing \( 1 – \cos x \). Use the identity \( 1 – \cos x = 2\sin^2\frac{x}{2} \).
Sandwich theorem g(x) is trapped between two functions whose limits are equal — then g has that same limit.
\( f'(a) = \lim_{h \to 0}\frac{f(a+h)-f(a)}{h} \) The question says ‘from first principle’, or asks for the derivative at a point using the definition. During the limit, h is not zero.
Product rule \( (uv)’ = u’v + uv’ \) Differentiating a product of two functions.
Quotient rule \( \left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2} \) Differentiating a quotient of two functions; the denominator is non-zero at the point.
Power rule and standard derivatives Differentiating \( x^n \), \( \sin x \), \( \cos x \), \( \tan x \), \( \cot x \) directly, without limits.

Worked Examples

Example 1: Evaluate the limit of sin 5x over 3x as x tends to 0

Step 1: Substituting x = 0 gives the 0/0 form, so use the standard limit \( \lim_{x \to 0}\frac{\sin x}{x} = 1 \).

The angle here is 5x, so match 5x in the denominator.

\[ \lim_{x \to 0}\frac{\sin 5x}{3x} = \lim_{x \to 0}\left( \frac{5}{3} \cdot \frac{\sin 5x}{5x} \right) \]

Step 2: Pull the constant out using the constant-multiple law and apply the sine limit with the angle 5x (as x tends to 0, 5x also tends to 0).

\[ = \frac{5}{3} \cdot \lim_{x \to 0}\frac{\sin 5x}{5x} = \frac{5}{3} \cdot 1 = \frac{5}{3} \]

Final answer: \( \frac{5}{3} \).

Example 2: Evaluate the limit of (x cubed minus 8) over (x minus 2) as x tends to 2

Step 1: Substituting x = 2 gives (8 minus 8) over (2 minus 2), the 0/0 form.

Recognise the standard pattern with a = 2 and n = 3.

Step 2: Apply the standard limit formula \( \lim_{x \to a}\frac{x^n – a^n}{x – a} = n a^{n-1} \):

\[ \lim_{x \to 2}\frac{x^3 – 8}{x – 2} = 3(2)^{3-1} = 3 \cdot 4 = 12 \]

Step 3 (check by factoring): \( x^3 – 8 = (x-2)(x^2+2x+4) \), so cancel \( x-2 \) (valid because x is not equal to 2 while taking the limit) and substitute:

\[ \lim_{x \to 2}(x^2 + 2x + 4) = 4 + 4 + 4 = 12 \]

Final answer: 12.

Example 3: Find the derivative of f(x) = 4x squared at x = 3 from first principles

Step 1: Write the first-principle definition at a = 3:

\[ f'(3) = \lim_{h \to 0}\frac{f(3+h) – f(3)}{h}, \quad f(3+h) = 4(3+h)^2, \quad f(3) = 36 \]

Step 2: Expand the numerator:

\[ = \lim_{h \to 0}\frac{4(9 + 6h + h^2) – 36}{h} = \lim_{h \to 0}\frac{24h + 4h^2}{h} \]

Step 3: During the limit h is not zero, so cancel h:

\[ = \lim_{h \to 0}(24 + 4h) = 24 \]

Final answer: \( f'(3) = 24 \), the slope of the tangent to \( y = 4x^2 \) at x = 3. Quick check: the power rule gives \( f'(x) = 8x \), so \( f'(3) = 8(3) = 24 \).

For limits and derivatives in any other chapter, the master maths formula index links every formula sheet on the site.

Common Mistakes to Avoid

These are the slips that show up most often when this chapter’s formulas are applied.

Mistake Correct rule How to check your answer
Assuming the limit must equal the function value at the point. The limit is decided by values of f near a; f(a) may differ from it or be undefined. NCERT Illustration 10: f(1) = 0, but the limit at 1 is 3.
Using the sine limit with the angle in degrees. \( \lim_{x \to 0}\frac{\sin x}{x} = 1 \) holds for radians. At x = 0.1 rad the ratio is about 0.998, close to 1; in degrees the ratio is about 0.017.
Stopping at 0/0 and writing ‘does not exist’. 0/0 means factor and cancel the common factor first; the limit may exist. \( \frac{x^2-4}{x-2} = x+2 \) for x not equal to 2, so the limit at 2 is 4.
Writing \( (uv)’ = u’v’ \). Product rule: \( (uv)’ = u’v + uv’ \). Put u = v = x: the rule must give 2x, not 1.
Reversing the numerator in the quotient rule. \( \left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2} \), not the other order. With u = 1, v = x the rule gives \( -\frac{1}{x^2} \), matching the known derivative of 1 over x.
Forgetting the minus sign in the derivative of cos x. \( \frac{d}{dx}(\cos x) = -\sin x \). cos x is decreasing near 0, so its slope there is negative.

Frequently Asked Questions

When does a limit not exist?

The common cases: the left hand limit and right hand limit differ, or one of them is not a finite value. For a rational function, if the denominator vanishes at a while the numerator does not, the limit does not exist. A 0/0 form is not automatically a non-existent limit — factor and cancel first.

Is the limit always equal to the value of the function?

No. The limit is fixed by values of f near a, not at a. It can exist when f(a) is undefined — see the rational function in Fig 12.2 — and it can disagree with the function value: in Fig 12.7, f(1) = 0 while the limit at 1 is 3.

What does ‘from first principle’ mean in derivative questions?

It means using the definition \( f'(x) = \lim_{h \to 0}\frac{f(x+h) – f(x)}{h} \) directly instead of the power, product or quotient rules. You substitute the given function, simplify the quotient, let h tend to 0, and read off the derivative.

Does the power rule work for negative and fractional powers?

The chapter proves \( \frac{d}{dx}(x^n) = n x^{n-1} \) for positive integers and notes the result holds for any real power.

The corresponding limit \( \lim_{x \to a}\frac{x^n – a^n}{x – a} = n a^{n-1} \) also holds for rational n when a is positive — this is how the NCERT example evaluates \( \lim_{x \to 0}\frac{\sqrt{1+x} – 1}{x} \).

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

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