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`        Could explain theses: Jump Disontinuity, Infinite Discontinuity, Oscillating Discontinuity???`
7 years ago

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:

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.
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.
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.
Example:
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.
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Siddharth !!!

Regards,
MOHIT
```
7 years ago
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