Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`
उत्तर
Given that `lim_(x -> 3) (x^2 - 9)/(x - 3)`
= `lim_(x -> 3) ((x + 3)(x - 3))/((x - 3))`
= `lim_(x -> 3) x + 3`
Taking limit, we have
3 + 3 = 6
Hence, the answer is 6.
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> 0) (sin ax + bx)/(ax + sin bx) a, b, a+ b != 0`
Evaluate the following limit.
`lim_(x -> 0) (cosec x - cot x)`
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 -> 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)]`
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
Find the positive integer n so that `lim_(x -> 3) (x^n - 3^n)/(x - 3)` = 108.
`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 -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`
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 -> 0) (1 - cos 2x)/x^2`
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 -> 0) (sin 2x + 3x)/(2x + tan 3x)`
Evaluate: `lim_(x -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
cos (x2 + 1)
`x^(2/3)`
Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> 0) |sinx|/x` is ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L 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 ______.