Advertisements
Advertisements
प्रश्न
`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`
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_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`
Evaluate the following limit :
`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`
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)]`
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 -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`
Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.
Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`
`lim_(x -> 0) |x|/x` is equal to ______.
Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
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) (2 sin x - sin 2x)/x^3`
Evaluate: `lim_(x -> 0) (1 - cos mx)/(1 - cos nx)`
Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`
`(ax + b)/(cx + d)`
`lim_(x -> pi) (1 - sin x/2)/(cos x/2 (cos x/4 - sin x/4))`
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.
`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is ______.
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
`lim_(x -> 3^+) x/([x])` = ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L is ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
