Advertisements
Advertisements
प्रश्न
Evaluate the following limit :
`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
उत्तर
`lim_(x -> pi/6) (2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)`
= `lim_(x -> pi/6) ((2sinx - 1)(sin x + 1))/((2sin x - 1)(sin x - 1))`
= `lim_(x -> pi/6) (sinx + 1)/(sin x - 1) ...[(because x -> pi/6"," x ≠ pi/6),(therefore sin x ≠ sin pi/6 = 1/2),(therefore 2 sin x - 1 ≠ 0)]`
= `(lim_(x -> pi/6) (sin x + 1))/(lim_(x -> pi/6) (sin x - 1))`
= `(sin pi/6 + 1)/(sin pi/6 - 1)`
= `(1/2 + 1)/(1/2 - 1)`
= `(1 + 2)/(1 - 2)`
= – 3
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit :
`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
Evaluate the following limit :
`lim_(x -> pi/6) [(2 - "cosec"x)/(cot^2x - 3)]`
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 → 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 -> "a") [(sinx - sin"a")/(x - "a")]`
`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 -> 0) (tanx - sinx)/(sin^3x)`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, 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 -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`
Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`
Evaluate: `lim_(x -> 0) (1 - cos 2x)/x^2`
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) (sqrt(3) sin x - cos x)/(x - pi/6)`
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 -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`
`lim_(x -> pi) (1 - sin x/2)/(cos x/2 (cos x/4 - sin x/4))`
`lim_(x -> pi) sinx/(x - pi)` is equal to ______.
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.
`lim_(x -> 0) (sin mx cot x/sqrt(3))` = 2, then m = ______.
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 ______.
The value of `lim_(x rightarrow 0) (4^x - 1)^3/(sin x^2/4 log(1 + 3x))`, is ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
