Advertisements
Advertisements
Question
`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.
Options
0
`1/2`
1
–1
Advertisements
Solution
`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to `1/2`.
Explanation:
We have `lim_(x -> 0) sinx/(x(1 + cos x))`
= `lim_(x -> 0) (2sin x/2 cos x/2)/(x 2 cos^2 x/2)`
= `1/2 lim_(x -> 0) (tan x/2)/(x/2)`
= `1/2`
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 x/(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) (cosec x - cot x)`
Evaluate the following limit :
`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
Evaluate the following limit :
`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`
Evaluate the following limit :
`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`
Evaluate the following limit :
`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
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 -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
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 -> 0) (sqrt(2 + x) - sqrt(2))/x`
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)`
`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 -> 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 2x + 3x)/(2x + tan 3x)`
Evaluate: `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
`x^(2/3)`
`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`
`lim_(x -> pi) sinx/(x - pi)` is equal to ______.
`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.
`lim_(x -> 0) (sin mx cot x/sqrt(3))` = 2, then m = ______.
The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
