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) cos x/(pi - x)`
Evaluate the following limit.
`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`
Evaluate the following limit.
`lim_(x -> (pi)/2) (tan 2x)/(x - pi/2)`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
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 -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`
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 -> 0) (tanx - sinx)/(sin^3x)`
`lim_(x -> 0) sinx/(x(1 + cos x))` is equal to ______.
`lim_(x -> pi/2) (1 - sin x)/cosx` 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 -> sqrt(2)) (x^4 - 4)/(x^2 + 3sqrt(2x) - 8)`
Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
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 -> 0) (sin 2x + 3x)/(2x + tan 3x)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
cos (x2 + 1)
x cos x
`lim_(x -> pi) (1 - sin x/2)/(cos x/2 (cos x/4 - sin x/4))`
Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists
`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.
`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.
`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
`lim_(x -> 0) (sin mx cot x/sqrt(3))` = 2, then m = ______.
`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 ______.
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 ______.
