Advertisements
Advertisements
Question
If `f(x) = {{:(sin[x]/[x]",", [x] ≠ 0),(0",", [x] = 0):}`, where [.] denotes the greatest integer function, then `lim_(x -> 0) f(x)` is equal to ______.
Options
1
0
– 1
None of these
Advertisements
Solution
If `f(x) = {{:((sin[x])/([x])",", [x] ≠ 0),(0",", [x] = 0):}`, where [.] denotes the greatest integer function, then `lim_(x -> 0) f(x)` is equal to none of these.
Explanation:
Given, `f(x) = {{:((sin[x])/([x])",", [x] ≠ 0),(0",", [x] = 0):}`
L.H.L = `lim_(x -> 0) (sin[x])/([x])`
= `lim_(h -> 0) (sin[0 - h])/([0 - h])`
= `lim_(h -> 0) (-sin[-h])/([-h])` = – 1
R.H.L = `lim_(x -> 0^+) (sin[x])/([x])`
= `lim_(h -> 0) (sin[0 + h])/([ 0 + h])`
= `lim_(h -> 0) (sin[h])/([h])` = 1
L.H.L ≠ R.H.L
So, the limit does not exist.
APPEARS IN
RELATED QUESTIONS
Evaluate the following limit.
`lim_(x -> pi) (sin(pi - x))/(pi (pi - x))`
Evaluate the following limit.
`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`
Evaluate the following limit.
`lim_(x -> 0) (cosec x - cot x)`
Evaluate the following limit :
`lim_(x -> 0) [(x*tanx)/(1 - cosx)]`
Evaluate the following limit :
`lim_(x -> pi/6) [(2 - "cosec"x)/(cot^2x - 3)]`
Evaluate the following limit :
`lim_(x -> pi/4) [(cosx - sinx)/(cos2x)]`
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)]`
Select the correct answer from the given alternatives.
`lim_(x -> 0) ((5sinx - xcosx)/(2tanx - 3x^2))` =
Evaluate the following :
`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`
`lim_{x→0}((3^x - 3^xcosx + cosx - 1)/(x^3))` is equal to ______
Evaluate `lim_(x -> 0) (sin(2 + x) - sin(2 - x))/x`
Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`
Find the derivative of f(x) = `sqrt(sinx)`, by first principle.
`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 -> a) ((2 + x)^(5/2) - (a + 2)^(5/2))/(x - a)`
Evaluate: `lim_(x -> 0) (sqrt(1 + x^3) - sqrt(1 - x^3))/x^2`
Evaluate: `lim_(x -> 0) (1 - cos mx)/(1 - cos nx)`
Evaluate: `lim_(x -> 0) (sqrt(2) - sqrt(1 + cos x))/(sin^2x)`
cos (x2 + 1)
`(ax + b)/(cx + d)`
`x^(2/3)`
`lim_(x -> 0) ((sin(alpha + beta) x + sin(alpha - beta)x + sin 2alpha x))/(cos 2betax - cos 2alphax) * 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))`
Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists
`lim_(x -> pi) sinx/(x - pi)` is equal to ______.
`lim_(x -> 0) ((1 + x)^n - 1)/x` is equal to ______.
`lim_(x -> pi/4) (sec^2x - 2)/(tan x - 1)` is equal to ______.
`lim_(x -> 0) |sinx|/x` is ______.
If `f(x) = {{:(x^2 - 1",", 0 < x < 2),(2x + 3",", 2 ≤ x < 3):}`, the quadratic equation whose roots are `lim_(x -> 2^-) f(x)` and `lim_(x -> 2^+) f(x)` is ______.
If L = `lim_(x→∞)(x^2sin 1/x - x)/(1 - |x|)`, then value of L is ______.
If `lim_(x→∞) 1/(x + 1) tan((πx + 1)/(2x + 2)) = a/(π - b)(a, b ∈ N)`; then the value of a + b is ______.
`lim_(x rightarrow π/2) ([1 - tan (x/2)] (1 - sin x))/([1 + tan (x/2)] (π - 2x)^3` is ______.
