मराठी

Show that limx→4|x-4|x-4 does not exists - Mathematics

Advertisements
Advertisements

प्रश्न

Show that `lim_(x -> 4) |x - 4|/(x - 4)` does not exists

बेरीज
Advertisements

उत्तर

Given `lim_(x -> 4) |x - 4|/(x - 4)` 

L.H.L. = `lim_(x -> 4^-) (-(x - 4))/(x - 4) = - 1`  ......`[because  |x - 4| = -(x - 4)  "if" x < 4]`

R.H.L. = `lim_(x -> 4^+) (x - 4)/(x - 4)` = 1   ......`[because |x - 4| = (x - 4)  "if"  x > 4]`

Since L.H.L. ≠ R.H.L.

Hence, the limit does not exist.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Limits and Derivatives - Exercise [पृष्ठ २४१]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
पाठ 13 Limits and Derivatives
Exercise | Q 51 | पृष्ठ २४१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate the following limit.

`lim_(x -> 0) (cos 2x -1)/(cos x - 1)`


Evaluate the following limit :

`lim_(theta -> 0) [(sin("m"theta))/(tan("n"theta))]`


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)]`


Select the correct answer from the given alternatives.

`lim_(x -> pi/2) [(3cos x + cos 3x)/(2x - pi)^3]` =


Evaluate the following :

`lim_(x -> 0) [(x(6^x - 3^x))/(cos (6x) - cos (4x))]`


Evaluate `lim_(x -> pi/6) (2sin^2x + sin x - 1)/(2sin^2 x - 3sin x + 1)`


Evaluate `lim_(x -> 0) (cos ax - cos bx)/(cos cx - 1)`


If f(x) = x sinx, then f" `pi/2` is equal to ______.


Evaluate: `lim_(x -> 3) (x^2 - 9)/(x - 3)`


Evaluate: `lim_(x -> 1/2) (4x^2 - 1)/(2x  - 1)`


Evaluate: `lim_(x -> 0) ((x + 2)^(1/3) - 2^(1/3))/x`


Evaluate: `lim_(x -> 1) (x^4 - sqrt(x))/(sqrt(x) - 1)`


Evaluate: `lim_(x -> sqrt(2)) (x^4 - 4)/(x^2 + 3sqrt(2x) - 8)`


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) (sin^2 2x)/(sin^2 4x)`


Evaluate: `lim_(x -> 0) (1 - cos mx)/(1 - cos nx)`


Evaluate: `lim_(x -> pi/4)  (sin x - cosx)/(x - pi/4)`


Evaluate: `lim_(x -> pi/6) (sqrt(3) sin x - cos x)/(x - pi/6)`


`lim_(y -> 0) ((x + y) sec(x + y) - x sec x)/y`


`lim_(x -> pi) sinx/(x - pi)` is equal to ______.


`lim_(x -> 1) (x^m - 1)/(x^n - 1)` 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 -> 1) ((sqrt(x) - 1)(2x - 3))/(2x^2 + x - 3)` is ______.


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 ______.


`lim_(x -> 0) |sinx|/x` is ______.


The value of `lim_(x → ∞) ((x^2 - 1)sin^2(πx))/(x^4 - 2x^3 + 2x - 1)` is equal to ______.


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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×