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 -> pi) (sin(pi - x))/(pi (pi - x))`
Evaluate the following limit.
`lim_(x -> 0) (ax + xcos x)/(b sin x)`
Evaluate the following limit.
`lim_(x -> 0) (cosec x - cot x)`
Evaluate the following limit :
`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`
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→-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)`
Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.
Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, is equal to ______.
If f(x) = x sinx, then f" `pi/2` is equal to ______.
Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
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 -> 0) (1 - cos 2x)/x^2`
Evaluate: `lim_(x -> 0) (2 sin x - sin 2x)/x^3`
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 -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`
`x^(2/3)`
`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`
`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`
Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` is ______.
`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
`lim_(x -> 3^+) x/([x])` = ______.
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 ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
