NCERT solutions for Mathematics Exemplar Class 11 chapter 13 - Limits and Derivatives [Latest edition]

Evaluate `lim_(x -> 2) 1/(x - 2) - (2(2x - 3))/(x^3 - 3x^2 + 2x)`

Evaluate `lim_(x -> 0) (sqrt(2 + x) - sqrt(2))/x`

Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.

Evaluate `lim_(x -> pi/2) (secx - tanx)`

Evaluate `lim_(x -> 0)  (sin(2 + x) - sin(2 - x))/x`

Find the derivative of f(x) = ax + b, where a and b are non-zero constants, by first principle

Find the derivative of f(x) = ax² + bx + c, where a, b and c are none-zero constant, by first principle.

Find the derivative of f(x) = x³, by first principle.

Find the derivative of f(x) = `1/x` by first principle.

Find the derivative of f(x) = sin x, by first principle.

Find the derivative of f(x) = xⁿ, where n is positive integer, by first principle.

Find the derivative of 2x⁴ + x.

Find the derivative of x² cosx.

Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`

Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`

Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`

Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`

Evaluate `lim_(h -> 0) ((a + h)^2 sin (a + h) - a^2 sina)/h`

Find the derivative of f(x) = tan(ax + b), by first principle.

Find the derivative of f(x) = `sqrt(sinx)`, by first principle.

Find the derivative of `cosx/(1 + sinx)`

`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.

`lim_(x -> pi/2) (1 - sin x)/cosx` is equal to ______.

`lim_(x -> 0) |x|/x` is equal to ______.

`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, is equal to ______.

`lim_(x -> 0) x sin  1/x` is equal to ______.

`lim_(n -> oo) (1 + 2 + 3 + ... + n)/n^2`, n ∈ N, is equal to ______.

If f(x) = x sinx, then f" `pi/2` is equal to ______.

Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`

Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x  - 1)`

Evaluate: `lim_(h -> 0) (sqrt(x + h) - sqrt(x))/h`

Evaluate: `lim_(x -> 0) ((x + 2)^(1/3) - 2^(1/3))/x`

Evaluate: `lim_(x -> 1) ((1 + x)^6 - 1)/((1 + x)^2 - 1)`

Evaluate: `lim_(x -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`

Evaluate: `lim_(x -> 1) (x^4 - sqrt(x))/(sqrt(x) - 1)`

Evaluate: `lim_(x -> 2) (x^2 - 4)/(sqrt(3x - 2) - sqrt(x + 2))`

Evaluate: `lim_(x -> sqrt(2)) (x^4 - 4)/(x^2 + 3sqrt(2x) - 8)`

Evaluate: `lim_(x -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`

Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`

Evaluate: `lim_(x -> 3) (x^3 + 27)/(x^5 + 243)`

Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`

Find ‘n’ if `lim_(x -> 2) (x^n - 2^n)/(x - 2)` = 80, x ∈ N

Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`

Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`

Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`

Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`

Evaluate: `lim_(x -> 0) (1 - cos mx)/(1 - cos nx)`

Evaluate: `lim_(x -> pi/3) (sqrt(1 - cos 6x))/(sqrt(2)(pi/3 - x))`

Evaluate: `lim_(x -> pi/4)  (sin x - cosx)/(x - pi/4)`

Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`

Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`

Evaluate: `lim_(x -> 0) (sin 2x + 3x)/(2x + tan 3x)`

Evaluate: `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`

Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec"  x - 2)`

Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`

Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`

If `lim_(x -> 1) (x^4 - 1)/(x - 1) = lim_(x -> k) (x^3 - l^3)/(x^2 - k^2)`, then find the value of k.

`(x^4 + x^3 + x^2 + 1)/x`

`(x + 1/x)^3`

(3x + 5)(1 + tan x)

(sec x – 1)(sec x + 1)

`(3x + 4)/(5x^2 - 7x + 9)`

`(x^5 - cosx)/sinx`

`(x^2 cos  pi/4)/sinx`

(ax² + cot x)(p + q cos x)

`(a + b sin x)/(c + d cos x)`

(sin x + cos x)²

(2x – 7)² (3x + 5)³ 

x² sin x + cos 2x

sin³x cos³x

`1/(ax^2 + bx + c)`

cos (x² + 1)

`(ax + b)/(cx + d)`

`x^(2/3)`

x cos x

`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`

`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`

`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`

`lim_(x -> pi) (1 - sin  x/2)/(cos  x/2 (cos  x/4 - sin  x/4))`

Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists

Let `f(x) = {{:((k cos x)/(pi - 2x)",", "when"  x ≠ pi/2),(3",", x = pi/2  "and if"  f(x) = f(pi/2)):}` find the value of k.

If `f(x) = {{:(x + 2",",  x ≤ - 1),(cx^2",", x > -1):}`, find 'c' if `lim_(x -> -1) f(x)` exists

`lim_(x -> pi) sinx/(x - pi)` is equal to ______.

`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.

`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.

`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.

`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is ______.

`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.

`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.

`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.

`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.

If `f(x) = {{:(sin[x]/[x]",", [x] ≠ 0),(0",", [x] = 0):}`, where [.] denotes the greatest integer function, then `lim_(x -> 0) f(x)` is equal to ______.

`lim_(x -> 0) |sinx|/x` is ______.

If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is ______. 

`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.

Let f(x) = x – [x]; ∈ R, then f'`(1/2)` is ______.

If `y = sqrt(x) + 1/sqrt(x)`, then`(dy)/(dx)` at x = 1 is  ______.

if `f(x) = (x - 4)/(2sqrt(x))`, then f'(1) is ______.

If `y = (1 + 1/x^2)/(1 - 1/x^2)` then `(dy)/(dx)` is ______.

If `y = (sin x + cos x)/(sin x - cos x)`, then `(dy)/(dx)` at x = 0 is ______.

If `y = (sin(x + 9))/cosx` then `(dy)/(dx)` at x = 0 is ______.

If `f(x) = 1 + x + x^2/2 + ... + x^100/100`, then f'(1) is equal to ______.

If `f(x) = (x^n - a^n)/(x - a)` for some constant, a, then f'(a) is equal to ______.

If `f(x) = x^100 + x^99 .... +  x + 1`, then f'(1) is equal to ______.

If `f(x) = 1 - x + x^2 - x^3 + ... -x^99 + x^100`, then f'(1) is equal to ______.

If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.

`lim_(x -> 0) (sin mx cot  x/sqrt(3))` = 2, then m = ______. 

If `y = 1 + x/(1!) + x^2/(2!) + x^3/(3!) + ...,` then `(dy)/(dx)` = ______.

`lim_(x -> 3^+) x/([x])` = ______.