Advertisements
Advertisements
प्रश्न
`lim_(x -> pi/2) (1 - sin x)/cosx` is equal to ______.
विकल्प
0
–1
1
Does not exit
उत्तर
`lim_(x -> pi/2) (1 - sin x)/cosx` is equal to 0.
Explanation:
`lim_(x -> pi/2) (1 - sin x)/cosx`
= `lim_(y -> 0) (1 - sin pi/2 - y)/(cos pi/2 - y)` taking ` pi/2 - x = y`
= `lim_(y -> 0) (1 - cos y)/siny`
= `lim_(y -> 0) (2 sin^2 t/2)/(2sin y/2 cos y/2)`
= `lim_(y -> 0) tan y/2`
= 0
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> (pi)/2) (tan 2x)/(x - pi/2)`
Evaluate the following limit :
`lim_(x ->0)((secx - 1)/x^2)`
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") [(sinx - sin"a")/(x - "a")]`
`lim_(x -> 0) sinx/(x(1 + cos x))` 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 -> 0) ((x + 2)^(1/3) - 2^(1/3))/x`
Evaluate: `lim_(x -> 1) (x^7 - 2x^5 + 1)/(x^3 - 3x^2 + 2)`
Evaluate: `lim_(x -> 1/2) (8x - 3)/(2x - 1) - (4x^2 + 1)/(4x^2 - 1)`
Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
Evaluate: `lim_(x -> pi/3) (sqrt(1 - cos 6x))/(sqrt(2)(pi/3 - x))`
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 -> a) (sin x - sin a)/(sqrt(x) - sqrt(a))`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
cos (x2 + 1)
`(ax + b)/(cx + d)`
`lim_(x -> pi/4) (tan^3x - tan x)/(cos(x + pi/4))`
`lim_(x -> pi) (1 - sin x/2)/(cos x/2 (cos x/4 - sin x/4))`
`lim_(x -> 0) (1 - cos 4theta)/(1 - cos 6theta)` 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) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.
`lim_(x rightarrow ∞) sum_(x = 1)^20 cos^(2n) (x - 10)` is equal to ______.
