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Siddharth M Grade: 12

Could explain theses: Jump Disontinuity, Infinite Discontinuity, Oscillating Discontinuity???

7 years ago

Answers : (1)

Askiitians Expert Mohit Singla
19 Points

Dear Siddharth,

Consider a function ƒ of real variable x with real values defined in a neighborhood of a point x0. Then three situations are possible in which the function ƒ is discontinuous at a point on the real axis x = x0:

  1. The limit from the negative direction
    and the  limit from the positive direction
    at x0 exist, are finite, and are equal. Then, if f(x0) is not equal to L = L + , x0 is called a removable discontinuity. This discontinuity can be removed to make f continuous at x0 by defining the value of function at x=x0.
  2. The limits L and L + exist and are finite, but not equal. Then, x0 is called a jump discontinuity. For this type of discontinuity, the value of f(x0) does not matter.
  3. One or both of the limits L and L + does not exist or is infinite. Then, x0 is called an infinite discontinuity.

An oscillating discontinuity occurs at a value of x near to which a function refuses to settle down.


The function y=sin(1/x) has a discontinuity at x = 0 because it is not defined at x = 0. It also has an oscillating discontinuity at x = 0,as here the left hand and right hand limit do not exist. 

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All the best Siddharth !!!



Askiitians Experts


7 years ago
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