Advertisements
Advertisements
प्रश्न
Evaluate the following :
`lim_(x -> 0)[(secx^2 - 1)/x^4]`
उत्तर
`lim_(x -> 0)(secx^2 - 1)/x^4`
Put x2 = y
As x → 0, x2 → 0
∴ y → 0
∴ Required limit
= `lim_(y -> 0) (sec y - 1)/(y^2)`
= `lim_(y -> 0) (sec y - 1)/(y^2) xx (sec y + 1)/(sec y + 1)`
= `lim_(y -> 0) (sec^2 y - 1)/(y^2(sec y + 1))`
= `lim_(y -> 0) (tan^2 y)/(y^2(sec y + 1))`
= `lim_(y -> 0) (tan^2 y)/(y^2) xx 1/(sec y + 1)`
= `lim_(y -> 0) ((tan^2 y)/y^2) xx lim_(y -> 0) 1/(sec y + 1)`
= `(1)^2 xx 1/(1 + 1)`
= `1/2`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`
Evaluate the following limit.
`lim_(x ->0) cos x/(pi - x)`
Evaluate the following limit.
`lim_(x -> 0) (sin ax + bx)/(ax + sin bx) a, b, a+ b != 0`
Evaluate the following limit.
`lim_(x -> (pi)/2) (tan 2x)/(x - pi/2)`
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)`
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
Evaluate the following :
`lim_(x -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`
Evaluate the following :
`lim_(x -> pi/4) [(sinx - cosx)^2/(sqrt(2) - sinx - cosx)]`
`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
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 ______.
`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 -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`
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 -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`
Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`
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)`
cos (x2 + 1)
`x^(2/3)`
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` 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) (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 ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
