मराठी

Evaluate limx→0cosax-cosbxcoscx-1

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

We have `lim_(x -> 0) (2sin  ((a + b))/2 x sin  ((a - b) x)/2)/(2 (sin^2  cx)/2)`

= `lim_(x -> 0) (2sin  ((a + b)x)/2 * sin  ((a - b)x)/2)/x^2 * x^2/(sin^2  (cx)/2)`

= `lim_(x -> 0) (sin  ((a + b)x)/2)/(((a + b)x)/2 * 2/(a + b)) * (sin   ((a - b)x)/2)/(((a - b)x)/2 * 2/(a - b)) * ((cx)^2/2 xx 4/c^2)/(sin^2  (cx)/2)`

= `(a + b)/2 xx (a - b)/2 xx 4/c^2`

= `(a^2 - b^2)/c^2`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Solved Examples [पृष्ठ २३४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 13 Limits and Derivatives
Solved Examples | Q 17 | पृष्ठ २३४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate the following limit.

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


Evaluate the following limit :

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


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)[(secx^2 - 1)/x^4]`


Evaluate the following :

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


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)`


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


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


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


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


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


cos (x2 + 1)


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


`x^(2/3)`


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


`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))`


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


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


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


`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 ______.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×