मराठी

Limx→1[x-1], where [.] is greatest integer function, is equal to ______.

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • 1

  • 2

  • 0

  • Does not exists

MCQ
रिकाम्या जागा भरा
Advertisements

उत्तर

`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, is equal to does not exists.

Explanation:

Since R.H.S. = `lim_(x -> 1^+) [x - 1]` = 0

And L.H.S. = `lim_(x -> 1^-) [x - 1]` = –1

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Solved Examples [पृष्ठ २३८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 13 Limits and Derivatives
Solved Examples | Q 25 | पृष्ठ २३८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate the following limit.

`lim_(x → 0) x sec x`


Evaluate the following limit.

`lim_(x -> 0) (sin ax + bx)/(ax + sin bx) a, b, a+ b != 0`


Evaluate the following limit.

`lim_(x -> (pi)/2) (tan 2x)/(x - pi/2)`


Evaluate the following limit :

`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`


Evaluate the following limit :

`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`


Evaluate the following limit :

`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`


Select the correct answer from the given alternatives.

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =


Evaluate the following :

`lim_(x -> "a") [(sinx - sin"a")/(x - "a")]`


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


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


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


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


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


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


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


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


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


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


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


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


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


`x^(2/3)`


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


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


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


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


If L = `lim_(x→∞)(x^2sin  1/x - x)/(1 - |x|)`, then value of L is ______.


If `lim_(x→∞) 1/(x + 1) tan((πx + 1)/(2x + 2)) = a/(π - b)(a, b ∈ N)`; then the value of a + b is ______.


If `lim_(n→∞)sum_(k = 2)^ncos^-1(1 + sqrt((k - 1)(k + 2)(k + 1)k)/(k(k + 1))) = π/λ`, then the value of λ is ______.


`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×