Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
Advertisements
उत्तर
Given that `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
= `lim_(x -> 0) (sin^2 2x)/(sin^2 2(2x))`
= `lim_(x -> 0) (sin^2 2x)/(4 sin^2 2x * cos^2 2x)` ......[sin 2x = 2 sin x cos x]
= `1/(4 cos^2 2x)`
Taking limit we have
= `1/(4 * cos^2 0)`
= `1/4`
APPEARS IN
संबंधित प्रश्न
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) (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_(x -> 0) [(x*tanx)/(1 - cosx)]`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
Evaluate the following limit :
`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`
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 → π/3) ((tan^2x - 3)/(sec^3x - 8))` =
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)`
Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`
`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.
`lim_(x -> pi/2) (1 - sin x)/cosx` is equal to ______.
`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 -> 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 -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`
Evaluate: `lim_(x -> 3) (x^3 + 27)/(x^5 + 243)`
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
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 -> 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)`
`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) sinx/(x - pi)` is equal to ______.
`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.
`lim_(x -> 0) |sinx|/x` is ______.
`lim_(x -> 3^+) x/([x])` = ______.
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 ______.
