Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> 0) (sin 3x)/(sin 7x)`
उत्तर
Given that `lim_(x -> 0) (sin 3x)/(sin 7x)`
= `lim_(x -> 0) ((sin 3x)/(3x) xx 3x)/((sin 7x)/(7x) xx 7x)`
= `(lim_(3x -> 0) ((sin 3x)/(3x)))/(lim_(7x -> 0) ((sin 7x)/(7x))) xx 3/7`
= `1/1 xx 3/7`
= `3/7` ......`[because lim_(x -> 0) sinx/x = 1]`
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`
Evaluate the following limit.
`lim_(x → 0) x sec 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/6) [(2 - "cosec"x)/(cot^2x - 3)]`
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)]`
Evaluate the following :
`lim_(x -> 0)[(secx^2 - 1)/x^4]`
Evaluate the following :
`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
Evaluate the following :
`lim_(x -> "a") [(x cos "a" - "a" cos x)/(x - "a")]`
Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`
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 -> 1) (x^4 - sqrt(x))/(sqrt(x) - 1)`
Evaluate: `lim_(x -> 0) (sin^2 2x)/(sin^2 4x)`
Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
cos (x2 + 1)
`lim_(x -> pi) (1 - sin x/2)/(cos x/2 (cos x/4 - sin x/4))`
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`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 ______.