Advertisements
Advertisements
Question
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
Advertisements
Solution
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
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) (ax + xcos x)/(b sin x)`
Evaluate the following limit.
`lim_(x → 0) x sec x`
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_(theta -> 0) [(1 - cos2theta)/theta^2]`
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/6) [(2 - "cosec"x)/(cot^2x - 3)]`
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 -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______
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) (sin(2 + x) - sin(2 - x))/x`
Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`
`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
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 -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`
Evaluate: `lim_(x -> 3) (x^3 + 27)/(x^5 + 243)`
Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`
x cos 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 - cos 4theta)/(1 - cos 6theta)` is ______.
`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
`lim_(x -> 0) |sinx|/x` is ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
`lim_(x -> 3^+) x/([x])` = ______.
Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L is ______.
The value of `lim_(x rightarrow 0) (4^x - 1)^3/(sin x^2/4 log(1 + 3x))`, is ______.
