Advertisements
Advertisements
प्रश्न
Evaluate the following :
`lim_(x -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`
Advertisements
उत्तर
`lim_(x -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`
= `lim_(x -> "a") [(x cos "a" - "a" cos "a" + "a" cos "a" - "a" cos x)/(x - "a")]` ...[Note this step]
= `lim_(x -> "a") [((x - "a") cos "a" + "a"(cos"a" - cosx))/(x - "a")]`
= `lim_(x -> "a") [((x - "a") cos "a" + 2"a" sin (("a" + x)/2)((x - "a")/2))/(x - "a")]`
= `lim_(x -> "a") [((x - "a")cos"a")/(x - "a") + (2"a" sin (("a" + x)/2) sin((x - "a")/2))/(x - "a")]`
= `lim_(x -> "a") [cos"a" + "a" sin (("a" + x)/2)* (sin((x - "a")/2))/(((x - "a")/2))]` ...[∵ x → a, x ≠ a, ∴ x – a ≠ 0]
= `lim_(x -> "a") cos"a" + "a"[lim_(x -> "a") sin(("a" + x)/2)] xx [lim_(x -> "a") sin((x - "a")/2)/((x - "a")/2)]`
= `cos "a" + "a" sin (("a" + "a")/2) xx 1 ...[(because x -> "a" "," x ≠ "a" therefore (x - "a")/2 -> 0),(and lim_(theta -> 0) sintheta/theta = 1)]`
= cos a + a sin a
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> 0) (sin ax + bx)/(ax + sin bx) a, b, a+ b != 0`
Evaluate the following limit :
`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`
Evaluate the following limit :
`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^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/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
Evaluate the following :
`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
Evaluate the following :
`lim_(x -> "a") [(sinx - sin"a")/(x - "a")]`
Evaluate `lim_(x -> 0) (sqrt(2 + x) - sqrt(2))/x`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
Evaluate: `lim_(x -> 1) (x^4 - sqrt(x))/(sqrt(x) - 1)`
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) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
`(ax + b)/(cx + d)`
`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 -> pi) sinx/(x - pi)` is equal to ______.
`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.
`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
`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 ______.
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.
`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 `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 ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
