Advertisements
Advertisements
प्रश्न
Evaluate the following limit.
`lim_(x → 0) x sec x`
Advertisements
उत्तर
`lim_(x → 0) x sec x`
= `lim_(x → 0)(x/cos x)`
= `0/cos 0`
= `0/1`
= 0
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
Evaluate the following limit :
`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`
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 -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =
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 ______
Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.
Evaluate `lim_(x -> pi/2) (secx - tanx)`
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`
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 -> 0) (1 - cos mx)/(1 - cos nx)`
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 -> 0) (sin 2x + 3x)/(2x + tan 3x)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
`x^(2/3)`
x cos x
`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.
`lim_(x -> 3^+) x/([x])` = ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L 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_(theta → -pi/4) (cos theta + sin theta)/(theta + pi/4)` =
