Advertisements
Advertisements
प्रश्न
Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
Advertisements
उत्तर
Given that `lim_(x -> 1/2) (4x^2 - 1)/(2x - 1)`
= `lim_(x -> 1/2) ((2x)^2 - (1)^2)/(2x - 1)`
= `lim_(x -> 1/2) ((2x + 1)(2x - 1))/(2x - 1)`
= `lim_(x -> 1/2) (2x + 1)`
Taking limit, we have
= `2 xx 1/2 + 1`
= 1 + 1
= 2.
APPEARS IN
संबंधित प्रश्न
Evaluate the following limit.
`lim_(x -> 0) (ax + xcos x)/(b sin 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 -> 0) (cosec x - cot x)`
Evaluate the following limit :
`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`
Evaluate the following limit :
`lim_(x -> pi/6) [(2 - "cosec"x)/(cot^2x - 3)]`
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
Evaluate the following :
`lim_(x -> 0)[(secx^2 - 1)/x^4]`
`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______
Evaluate `lim_(x -> 2) 1/(x - 2) - (2(2x - 3))/(x^3 - 3x^2 + 2x)`
Evaluate `lim_(x -> 0) (sin(2 + x) - sin(2 - x))/x`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
`lim_(x -> 1) [x - 1]`, where [.] is greatest integer function, is equal to ______.
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 -> 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 2x)/x^2`
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 -> pi/6) (cot^2 x - 3)/("cosec" x - 2)`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
`x^(2/3)`
x cos x
`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 -> pi) sinx/(x - pi)` is equal to ______.
`lim_(x -> 0) (x^2 cosx)/(1 - cosx)` is ______.
`lim_(x -> 0) ("cosec" x - cot x)/x` is equal to ______.
`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
`lim_(x -> 0) |sinx|/x` is ______.
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.
The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.
