Advertisements
Advertisements
प्रश्न
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to ______.
विकल्प
2
`1/2`
`-1/2`
`1/4`
Advertisements
उत्तर
`lim_(x -> 0) (tan 2x - x)/(3x - sin x)` is equal to `1/2`.
Explanation:
Given `lim_(x -> 0) (tan 2x - x)/(3x - sin x)`
= `lim_(x -> 0) (x[tan2x/x - 1])/(x[3 - sin x/x])`
`lim_((x -> 0),(because 2x -> 0)) ((tan 2x)/(2x) xx 2 - 1)/(3 - sinx/x)`
= `(1.2 - 1)/(3 - 1)`
= `(2 - 1)/2`
= `1/2`
∴ 2x → 0
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) (cosec x - cot x)`
Evaluate the following limit :
`lim_(theta -> 0) [(1 - cos2theta)/theta^2]`
Evaluate the following limit :
`lim_(x -> 0)[(1 - cos("n"x))/(1 - cos("m"x))]`
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)]`
Evaluate the following limit :
`lim_(x -> pi/6) [(2sin^2x + sinx - 1)/(2sin^2x - 3sinx + 1)]`
Select the correct answer from the given alternatives.
`lim_(x → π/3) ((tan^2x - 3)/(sec^3x - 8))` =
Select the correct answer from the given alternatives.
`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =
`lim_{x→-5} (sin^-1(x + 5))/(x^2 + 5x)` is equal to ______
Evaluate `lim_(x -> 0) (sin(2 + x) - sin(2 - x))/x`
Evaluate `lim_(x -> 0) (tanx - sinx)/(sin^3x)`
Evaluate `lim_(x -> a) (sqrt(a + 2x) - sqrt(3x))/(sqrt(3a + x) - 2sqrt(x))`
Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
`lim_(x -> 0) sinx/(x(1 + cos x))` 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) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
Evaluate: `lim_(x -> 0) (sin x - 2 sin 3x + sin 5x)/x`
cos (x2 + 1)
`(ax + b)/(cx + d)`
`lim_(x -> pi) (1 - sin x/2)/(cos x/2 (cos x/4 - sin x/4))`
Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists
`lim_(x -> 1) (x^m - 1)/(x^n - 1)` is ______.
`lim_(x -> 0) sinx/(sqrt(x + 1) - sqrt(1 - x)` is ______.
If `f(x) = tanx/(x - pi)`, then `lim_(x -> pi) f(x)` = ______.
Let Sk = `sum_(r = 1)^k tan^-1(6^r/(2^(2r + 1) + 3^(2r + 1)))`. Then `lim_(k→∞)` Sk = is equal to ______.
The value of `lim_(x rightarrow 0) (4^x - 1)^3/(sin x^2/4 log(1 + 3x))`, is ______.
