Advertisements
Advertisements
प्रश्न
`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`
Advertisements
उत्तर
Given, `lim_(x -> 0) ([sin (alpha + beta)x + sin(alpha - beta)x + sin 2alpha * x])/(cos 2betax - cos 2alphax) * x`
= `lim_(x -> 0) ([2 sin alpha x * cos betax + sin 2alpha * x]*x)/(2sin (alpha + beta)x * sin(alpha - beta)x)` ......`[(because sin C + sin D = 2 sin (C + D)/2 * cos (C - D)/2),(cos C - cos D = - 2 sin (C + D)/2 * sin (C - D)/2)]`
= `lim_(x -> 0) ([2 sin alphax * cos betax + 2 sin alphax * cos alphax] * x)/(2 sin(alpha + beta)x * sin(alpha - beta)x)`
= `lim_(x -> 0) (2 sin alphax (cos betax + cos alphax)*x)/(2 sin(alpha + beta)x * sin(alpha - beta)x)`
= `lim_(x -> 0) (sin alphax[2 cos((alpha + beta)/2)x * cos((alpha - beta)/2)x]*x)/(sin(alpha + beta)x * sin(alpha - beta)x)`
= `lim_(x -> 0) (sin alphax [2 cos ((alpha + beta)/2)x * cos((alpha - beta)/2)x]*x)/(2 sin((alpha + beta)/2)x * cos((alpha + beta)/2)x) * 2 sin ((alpha - beta)/2)x * cos((alpha - beta)/2)x` .......`[(because cos C + cos D = 2 cos (C + d)/2 cos (C - D)/2),("and" sin 2x = 2 sin x cos x)]`
= `lim_(x -> 0) (sin alphax * x)/(2sin((alpha + beta)/2)x sin((alpha - beta)/2) * x)`
= `lim_(x -> 0) 1/2 ((sin alphax)/(alphax) * (alphax) * x)/([(sin ((alpha + beta)/2) x)/(((alpha + beta)/2) * x) xx ((alpha + beta)/2) * x][(sin((alpha - beta)/2)*x)/(((alpha - beta)/2)* x) xx ((alpha - beta))/2 * x])`
= `1/2 * (alphax^2)/(((alpha + beta)/2)x * ((alpha - beta)/2)x)`
= `1/2[alpha/(((alpha + beta)/2)((alpha - beta)/2))]`
= `1/2 * (4alpha)/(alpha^2 - beta^2)`
= `(2alpha)/(alpha^2 - beta^2)`
APPEARS IN
संबंधित प्रश्न
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)((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 -> 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)]`
Evaluate the following limit :
`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"x)]`
Evaluate the following limit :
`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
Select the correct answer from the given alternatives.
`lim_(x → π/3) ((tan^2x - 3)/(sec^3x - 8))` =
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^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 -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`
If f(x) = x sinx, then f" `pi/2` is equal to ______.
Evaluate: `lim_(x -> sqrt(2)) (x^4 - 4)/(x^2 + 3sqrt(2x) - 8)`
Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`
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/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 -> 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)`
`(ax + b)/(cx + d)`
x cos x
`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`
`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`
Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists
`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
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) (tan 2x - x)/(3x - sin 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 ______.
Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L is ______.
