Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
Advertisements
उत्तर
Given that `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
= `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x) xx (sqrt(2) + sqrt(1 + cosx))/(sqrt(2) + sqrt(1 + cos x))`
= `lim_(x -> 0) (2 - (1 + cos x))/(sin^2x [sqrt(2) + sqrt(1 + cos x)])`
= `lim_(x -> 0) (1 - cos x)/(sin^2 xx [sqrt(2) + sqrt(1 + cos x)])`
= `lim_(x -> 0) (2 sin^2 x/2)/((2 sin x/2 cos x/2)^2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`
= `lim_(x -> 0) (2 sin^2 x/2)/(4 sin^2 x/2 cos^2 x/2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`
= `lim_(x -> 0) 2/(4 cos^2 x/2) xx 1/([sqrt(2) + sqrt(1 + cos x)])`
Taking limit, we get
= `2/(4 cos^2 0) xx 1/((sqrt(2) + sqrt(2))`
= `1/2 xx 1/(2sqrt(2))`
= `1/(4sqrt(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 -> 0) (cosec x - cot x)`
Evaluate the following limit :
`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`
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) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`
Evaluate the following limit :
`lim_(x -> pi/4) [(tan^2x - cot^2x)/(secx - "cosec"x)]`
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) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
Evaluate the following :
`lim_(x -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`
Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.
Evaluate `lim_(x -> pi/2) (secx - tanx)`
Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 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 -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`
Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`
Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`
Evaluate: `lim_(x -> pi/3) (sqrt(1 - cos 6x))/(sqrt(2)(pi/3 - x))`
Evaluate: `lim_(x -> pi/4) (sin x - cosx)/(x - pi/4)`
Evaluate: `lim_(x -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
x cos x
`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * x`
`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` 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)` = ______.
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 `lim_(x→∞) 1/(x + 1) tan((πx + 1)/(2x + 2)) = a/(π - b)(a, b ∈ N)`; then the value of a + b is ______.
If `lim_(n→∞)sum_(k = 2)^ncos^-1(1 + sqrt((k - 1)(k + 2)(k + 1)k)/(k(k + 1))) = π/λ`, then the value of λ is ______.
