Advertisements
Advertisements
प्रश्न
Evaluate the following limit :
`lim_(x -> 0)[(1 - cos("n"x))/(1 - cos("m"x))]`
उत्तर
`lim_(x -> 0)(1 - cos"n"x)/(1 - cos"m"x)`
= `lim_(x -> 0) (1 - cos"n"x)/(1 - cos"m"x) xx (1 + cos "n"x)/(1 + cos "n"x) xx (1 + cos"m"x)/(1 + cos"m"x)`
= `lim_(x -> 0) ((1 - cos^2"n"x)(1 + cos"m"x))/((1 - cos^2"m"x)(1 + cos "n"x))`
= `lim_(x -> 0) (sin^2"n"x(1 + cos "m"x))/(sin^2"m"x(1 + cos "n"x))`
= `lim_(x -> 0) (((sin^2"n"x)/("n"^2x^2))(1 + cos "m"x))/(((sin^2"m"x)/("m"^2x^2))(1 + cos "n"x)) xx "n"^2/"m"^2` ...[∵ x → 0, x ≠ 0 ∴ x2 ≠ 0]
= `"n"^2/"m"^2 ([lim_(x -> 0) (sin"n"x)/("n"x)]^2 xx [lim_(x -> 0) (1 + cos "m"x)])/([lim_(x -> 0) (sin"m"x)/("m"x)]^2 xx [lim_(x -> 0) (1 + cos "n"x)])`
= `"n"^2/"m"^2 (1^2*(1 + cos 0))/(1^2*(1 + cos 0)) ...[because x -> 0 therefore "m"x, "n"x -> 0 and lim_(theta -> 0) sintheta/theta = 1]`
= `"n"^2/"m"^2`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> 0) (cosec x - cot x)`
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) [(x*tanx)/(1 - cosx)]`
Evaluate the following limit :
`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`
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))]`
Evaluate the following :
`lim_(x -> "a") [(x cos "a" - "a" cos x)/(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)`
Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.
Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`
`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.
`lim_(x -> 0) |x|/x` is equal to ______.
`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, is equal to ______.
Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`
Evaluate: `lim_(x -> 0) ((x + 2)^(1/3) - 2^(1/3))/x`
Evaluate: `lim_(x -> sqrt(2)) (x^4 - 4)/(x^2 + 3sqrt(2x) - 8)`
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
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)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`
`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) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.
`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.
`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
`lim_(x -> 0) (sin mx cot x/sqrt(3))` = 2, then m = ______.
`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 ______.
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 ______.
