Advertisements
Advertisements
प्रश्न
Evaluate the following :
`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`
Advertisements
उत्तर
`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`
= `lim_(x -> pi/4) (1 - sin2x)/(sqrt(2) - (sin x + cos x))`
= `lim_(x -> pi/4) (1 - sin2x)/(sqrt(2) - sqrt(1 + sin2))`
Put 1 + sin 2x = t
∴ sin 2x = t – 1
As `x -> pi/4, "t" -> 1 + sin 2 (pi/4)`
∴ `"t" -> 1 + sin pi/2`
∴ t → 1 + 1
∴ t → 2
∴ Required limit
= `lim_("t" -> 2) (1 - ("t" - 1))/(sqrt(2) - sqrt("t"))`
= `lim_("t" -> 2) (2 - "t")/(2^(1/2) - "t"^(1/2))`
= `lim_("t" -> 2) ("t" - 2)/("t"^(1/2) - 2^(1/2))`
= `lim_("t" -> 2) 1/(("t"^(1/2) - 2^(1/2))/("t" - 2)) ...[("Divide Numerator and Denominator"),("by" "t" - 2 "As" "t" -> 2 "," "t" ≠ 2),(therefore "t" - 2 ≠ 0)]`
= `1/(1/2(2)^((-1)/2)`
= `2(2)^(1/2)`
= `2sqrt(2)`
APPEARS IN
संबंधित प्रश्न
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_(x -> 0) (cosec x - cot x)`
Evaluate the following limit :
`lim_(theta -> 0) [(1 - cos2theta)/theta^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)]`
Evaluate the following limit :
`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"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))` =
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
Evaluate the following :
`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______
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)`
`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.
Evaluate: `lim_(x -> 0) ((x + 2)^(1/3) - 2^(1/3))/x`
Evaluate: `lim_(x -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`
Evaluate: `lim_(x -> 1) (x^4 - sqrt(x))/(sqrt(x) - 1)`
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 -> 0) (sin 3x)/(sin 7x)`
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
Evaluate: `lim_(x -> 0) (1 - cos mx)/(1 - cos nx)`
Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
`(ax + b)/(cx + d)`
`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.
`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 ______.
