हिंदी

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 ______.

विकल्प

  • 1

  • 2

  • 0

  • Does not exists

MCQ
रिक्त स्थान भरें
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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 13: Limits and Derivatives - Solved Examples [पृष्ठ २३८]

APPEARS IN

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

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Evaluate the following limit.

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


Evaluate the following limit.

`lim_(x → 0) x sec x`


Evaluate the following limit :

`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`


Evaluate the following limit :

`lim_(x ->0)((secx - 1)/x^2)`


Evaluate the following limit :

`lim_(x -> 0)[(1 - cos("n"x))/(1 - cos("m"x))]`


Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`


Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`


Select the correct answer from the given alternatives.

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


Evaluate the following :

`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`


`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______ 


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


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


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


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


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


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 -> 0) (sin 3x)/(sin 7x)`


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


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


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


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


cos (x2 + 1)


`x^(2/3)`


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


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


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


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


The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.


Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×