Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`
Advertisements
उत्तर
Given that `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`
= `lim_(x -> pi/6) (2[sqrt(3)/2 sin x - 1/2 cos x])/(x - pi/6)`
= `lim_(x -> pi/6) (2[cos pi/6 sin x - sin pi/6 cos x])/(x - pi/6)`
= `lim_((x -> pi/6),(because x - pi/6 -> 0)) (2sin (x - pi/6))/((x - pi/6))` .......`[because lim_(x -> 0) sinx.x = 1]`
= `2 * 1`
= 2
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`
Evaluate the following limit.
`lim_(x -> 0) (ax + xcos x)/(b sin x)`
Evaluate the following limit :
`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`
Evaluate the following limit :
`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`
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 → π/3) ((tan^2x - 3)/(sec^3x - 8))` =
Select the correct answer from the given alternatives.
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
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 -> pi/2) (1 - sin x)/cosx` is equal to ______.
`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, is equal to ______.
Evaluate: `lim_(x -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`
Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`
Evaluate: `lim_(x -> 3) (x^3 + 27)/(x^5 + 243)`
Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`
Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
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 -> 0) (sin 2x + 3x)/(2x + tan 3x)`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
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 -> 1) (x^m - 1)/(x^n - 1)` 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) (tan 2x - x)/(3x - sin x)` is equal to ______.
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_(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 ______.
