Advertisements
Advertisements
प्रश्न
Evaluate the following limit :
`lim_(x -> pi) [(sqrt(1 - cosx) - sqrt(2))/(sin^2 x)]`
Advertisements
उत्तर
`lim_(x -> pi) (sqrt(1 - cosx) - sqrt(2))/(sin^2 x)`
= `lim_(x -> pi) (sqrt(1 - cosx) - sqrt(2))/(sin^2x) xx (sqrt(1 - cosx) + sqrt(2))/(sqrt(1 - cosx) + sqrt(2))`
= `lim_(x -> pi) ((1 - cos x) - 2)/(( 1 - cos^2x)(sqrt(1 - cos x) + sqrt(2))`
= `lim_(x -> pi) (-(1 + cos x))/((1 + cos x)(1 - cosx)(sqrt(1 - cosx) + sqrt(2))`
= `lim_(x -> pi) (-1)/((1 - cosx)(sqrt(1 - cosx) + sqrt(2))) ...[(because x -> pi"," x ≠ pi),(therefore cos x ≠ cos pi = -1),(therefore 1 + cos x ≠ 0)]`
= `(lim_(x -> pi) (-1))/([lim_(x -> pi) (1 - cosx)] xx [lim_(x -> pi) (sqrt(1 - cosx) + sqrt(2))]`
= `(-1)/((1 - cos pi) (sqrt(1 - cos pi) + sqrt(2))`
= `(-1)/((1 + 1)(sqrt(1 + 1) + sqrt(2))`
= `(-1)/(4sqrt(2))`.
APPEARS IN
संबंधित प्रश्न
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 -> (pi)/2) (tan 2x)/(x - pi/2)`
Evaluate the following limit :
`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`
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))` =
`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______
Evaluate `lim_(x -> 2) 1/(x - 2) - (2(2x - 3))/(x^3 - 3x^2 + 2x)`
Evaluate `lim_(x -> 0) (sqrt(2 + x) - sqrt(2))/x`
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) sinx/(x(1 + cos x))` is equal to ______.
`lim_(x -> pi/2) (1 - sin x)/cosx` is equal to ______.
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
Evaluate: `lim_(x -> 1) (x^4 - sqrt(x))/(sqrt(x) - 1)`
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) (1 - cos mx)/(1 - cos nx)`
Evaluate: `lim_(x -> pi/4) (sin x - cosx)/(x - pi/4)`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
`(ax + b)/(cx + d)`
`lim_(x -> pi) sinx/(x - pi)` is equal to ______.
`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.
`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.
`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is ______.
`lim_(x -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.
If `f(x) = {{:(sin[x]/[x]",", [x] ≠ 0),(0",", [x] = 0):}`, where [.] denotes the greatest integer function, then `lim_(x -> 0) f(x)` is equal to ______.
`lim_(x -> 0) |sinx|/x` is ______.
If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L is ______.
`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.
