Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
Advertisements
उत्तर
Given that `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
= `lim_(x -> 0) (2 sin x - 2 sin x cos x)/x^3`
= `lim_(x -> 0) (2 sin x(1 - cosx))/x^3`
= `lim_(x -> 0) (2sinx)x (( - cosx)/x)`
= `lim_(x -> ) ((sinx)/x)((sin^2 x/2)/x^2)`
= `lim_(x -> 0) 2((sinx)/x)(2 (sin^2 x/2)/(x^2/4) xx 1/4)`
= `lim_(x -> 0) 2((sin x)/x) 2[((sin x/2)/(x/2))^2] * 1/4`
= `lim_(x -> 0) 4/4 ((sin x)x)`
= `lim_(x/2 -> 0) ((sin x/2)/(x/2))^2`
= `1 * 1 * (1)^2`
= 1 .....`[because lim_(x -> 0) sinx/x = 1]`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`
Evaluate the following limit.
`lim_(x -> 0) (ax + xcos x)/(b sin x)`
Evaluate the following limit.
`lim_(x → 0) x sec 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_(theta -> 0) [(1 - cos2theta)/theta^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)]`
Select the correct answer from the given alternatives.
`lim_(x → π/3) ((tan^2x - 3)/(sec^3x - 8))` =
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
Select the correct answer from the given alternatives.
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =
Evaluate the following :
`lim_(x -> 0)[(secx^2 - 1)/x^4]`
Evaluate the following :
`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`
`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______
Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.
Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
`lim_(x -> 0) |x|/x` is equal to ______.
Evaluate: `lim_(x -> 0) ((x + 2)^(1/3) - 2^(1/3))/x`
Evaluate: `lim_(x -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`
Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`
Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`
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)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
cos (x2 + 1)
`(ax + b)/(cx + d)`
`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
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_(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 ______.
