Advertisements
Advertisements
प्रश्न
Evaluate the following limit :
`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`
Advertisements
उत्तर
`lim_(x -> 0) (cos("a"x) - cos("b"x))/(cos("c"x) - 1)`
= `lim_(x -> 0) (-cos("a"x) + cos("b"x))/(1 - cos("c"x))`
= `lim_(x -> 0) (1 - cos "a"x - 1 + cos "b"x)/(1 - cos "c"x)`
= `lim_(x -> 0) ((1 - cos"a"x) - (1 - cos"b"x))/(1 - cos"c"x)`
= `lim_(x -> 0) (2sin^2 (("a"x)/2) - 2sin^2 ("b"x)/2)/(2sin^2 ("c"x)/2`
= `lim_(x -> 0) ((sin^2 (("a"x)/2) - sin^2 (("b"x)/2))/(x^2))/((sin^2 (("c"x)/2))/(x^2)) ...[("Divide numerator and denominator by" x^2),(because x -> 0"," therefore x ≠ 0 therefore x^2 ≠ 0)]`
= `(lim_(x -> 0) [(sin^2 (("a"x)/2))/x^2 - (sin^2(("b"x)/2))/x^2])/(lim_(x -> 0) (sin^2 (("c"x)/2))/x^2)`
= `(lim_(x -> 0) [(sin ("a"x)/2)/x]^2 - lim_(x -> 0) [(sin ("b"x)/2)/x]^2)/(lim_(x -> 0) [[sin ("c"x)/2)/x]^2`
= `(lim_(x -> 0) [(sin ("a"x)/2)/(("a"x)/2)]^2 * ("a"/2)^2 - lim_(x -> 0) [(sin ("b"x)/2)/(("b"x)/2)]^2 * ("b"/2)^2)/(lim_(x -> 0) [(sin ("c"x)/2)/(("c"x)/2)]^2 * ("c"/2)^2`
= `((1)^2 * "a"^2/4 - (1)^2 * "b"^2/4)/((1)^2 * "c"^2/4`
= `("a"^2/4 - "b"^2/4)/("c"^2/4)`
= `("a"^2 - "b"^2)/"c"^2`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`
Evaluate the following limit.
`lim_(x ->0) cos x/(pi - x)`
Evaluate the following limit.
`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`
Evaluate the following limit.
`lim_(x -> 0) (sin ax + bx)/(ax + sin bx) a, b, a+ b != 0`
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 ->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 -> 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)]`
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 -> pi/2) (secx - tanx)`
Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
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/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)`
`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 + x)^n - 1)/x` is equal to ______.
`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.
`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.
`lim_(x -> 0) |sinx|/x` 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) (tan 2x - x)/(3x - sin x)` is equal to ______.
`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
`lim_(theta → -pi/4) (cos theta + sin theta)/(theta + pi/4)` =
