Advertisements
Advertisements
Question
`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`
Solution
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
RELATED QUESTIONS
Evaluate the following limit.
`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`
Evaluate the following limit.
`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`
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 :
`lim_(x -> 0)[(secx^2 - 1)/x^4]`
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)]`
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) sinx/(x(1 + cos x))` is equal to ______.
Evaluate: `lim_(x -> sqrt(2)) (x^4 - 4)/(x^2 + 3sqrt(2x) - 8)`
Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`
Evaluate: `lim_(x -> pi/4) (sin x - cosx)/(x - pi/4)`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
`x^(2/3)`
x cos x
`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`
`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 ______.
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 -> 0) (sin mx cot x/sqrt(3))` = 2, then m = ______.
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 `lim_(x→∞) 1/(x + 1) tan((πx + 1)/(2x + 2)) = a/(π - b)(a, b ∈ N)`; then the value of a + b is ______.
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 ______.
The value of `lim_(x rightarrow 0) (4^x - 1)^3/(sin x^2/4 log(1 + 3x))`, is ______.
