Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`
उत्तर
Given that `lim_(x -> 0) (1 - cos 2x)/x^2`
= `lim_(x -> 0) (2 sin^2x)/x^2` .......[cos 2x = 1 – 2 sin2x]
= `lim_(x -> 0) 2(sin x/x)^2`
= `2 xx 1`
= 2 .....`[because lim_(x -> 0) sinx/x = 1]`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x ->0) cos x/(pi - x)`
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 -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
Evaluate the following :
`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
Evaluate the following :
`lim_(x -> "a") [(sinx - sin"a")/(x - "a")]`
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`
`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, is equal to ______.
If f(x) = x sinx, then f" `pi/2` is equal to ______.
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`
Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
Evaluate: `lim_(x -> 0) (1 - cos mx)/(1 - cos nx)`
Evaluate: `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
`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 -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` 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 ______.
`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 ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.
`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.