Advertisements
Advertisements
Question
Evaluate the following limit :
`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`
Advertisements
Solution
`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
RELATED QUESTIONS
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_(x -> 0) (cosec x - cot x)`
Evaluate the following limit :
`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`
Evaluate the following limit :
`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
Evaluate the following limit :
`lim_(x -> pi/6) [(2 - "cosec"x)/(cot^2x - 3)]`
Evaluate the following :
`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
Evaluate `lim_(x -> 2) 1/(x - 2) - (2(2x - 3))/(x^3 - 3x^2 + 2x)`
Evaluate `lim_(x -> 0) (sin(2 + x) - sin(2 - x))/x`
`lim_(x -> 0) |x|/x` is equal to ______.
Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`
Evaluate: `lim_(x -> 1) (x^4 - sqrt(x))/(sqrt(x) - 1)`
Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`
Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`
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/6) (sqrt(3) sin x - cos x)/(x - pi/6)`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
`(ax + b)/(cx + d)`
`x^(2/3)`
`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))`
`lim_(x -> pi) (1 - sin x/2)/(cos x/2 (cos x/4 - sin x/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 -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
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 ______.
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 ______.
The value of `lim_(x rightarrow 0) (4^x - 1)^3/(sin x^2/4 log(1 + 3x))`, is ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
