हिंदी

Select the correct answer from the given alternatives. limx→0(5sinx-xcosx2tanx-3x2) = - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Select the correct answer from the given alternatives.

`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =

विकल्प

  • 0

  • 1

  • 2

  • 3

MCQ

उत्तर

2

Explanation;

`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` 

= `(lim_(x -> 0) ((5sinx)/x - cosx))/(lim_(x -> 0) ((2tanx)/x - 3x))`

= `(5(1) - cos0)/(2(1) - 3(0))`

= 2

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Limits - Miscellaneous Exercise 7.1 [पृष्ठ १५८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 7 Limits
Miscellaneous Exercise 7.1 | Q I. (6) | पृष्ठ १५८

संबंधित प्रश्न

Evaluate the following limit.

`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`


Evaluate the following limit.

`lim_(x ->0) cos x/(pi - x)`


Evaluate the following limit.

`lim_(x -> 0) (ax +  xcos x)/(b sin x)`


Evaluate the following limit.

`lim_(x → 0) x sec x`


Evaluate the following limit.

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


Evaluate the following limit :

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


Evaluate the following limit :

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


Evaluate the following limit :

`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"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))]`


Evaluate the following :

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


Evaluate the following :

`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`


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


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


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


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


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


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


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


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


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))`


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


`x^(2/3)`


x cos x


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


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


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


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


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


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


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


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 π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×