Advertisements
Advertisements
प्रश्न
Select the correct answer from the given alternatives.
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =
विकल्प
`3/2`
`1/2`
`-1/2`
`1/4`
उत्तर
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3] = underline (-1/2)`
Explanation:
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]`
= `lim_(x -> pi/2)(3cos x + 4cos^3x - 3cos x)/(2x - pi)^3`
= `lim_(x -> pi/2) (4cos^3x) /(8(x - pi/2)^3`
Put `x = pi/2 + h,`
`x - pi/2 = h`
As `x -> pi/2, h -> 0`
= `lim_(x -> pi/2) (4 cos^3x) /(8(x - pi/2)^3`
= `lim_(h-> 0) (4 cos^3(pi/2 + h))/(8h^3)`
= `lim_(h -> 0) (4(-sin h)^3)/(8h^3)` ...`[∵ cos(pi/2 + θ) = -sinθ]`
= `-1/2(lim_(h->0)(sin h)/h)^3`
= `-1/2`
APPEARS IN
संबंधित प्रश्न
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 -> (pi)/2) (tan 2x)/(x - pi/2)`
Evaluate the following limit :
`lim_(x -> pi/6) [(2 - "cosec"x)/(cot^2x - 3)]`
Evaluate the following limit :
`lim_(x -> 0) [(cos("a"x) - cos("b"x))/(cos("c"x) - 1)]`
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
Evaluate the following :
`lim_(x -> "a") [(sinx - sin"a")/(x - "a")]`
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
Evaluate `lim_(x -> 2) 1/(x - 2) - (2(2x - 3))/(x^3 - 3x^2 + 2x)`
Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`
Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
`lim_(x -> 0) |x|/x` 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 -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`
Evaluate: `lim_(x -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`
Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
cos (x2 + 1)
`(ax + b)/(cx + d)`
`x^(2/3)`
`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`
Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists
`lim_(x -> pi) sinx/(x - pi)` is equal to ______.
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
`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 ______.
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 ______.
If `lim_(x→∞) 1/(x + 1) tan((πx + 1)/(2x + 2)) = a/(π - b)(a, b ∈ N)`; then the value of a + b is ______.
`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.
