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CONTINUITY – ALL YOU NEED TO KNOW

We know that for every point, there exists neighboring points.
But when we say that the function f(x) is continuous at a point x=a, we mean that at point ( a, f(a) ) the graph of the function f(x) has no holes or gaps or discontinuity. That is, its graph is unbroken at point (a, f(a))


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We know that for every point, there exists neighboring points.

But when we say that the function f(x) is continuous at a point x=a, we mean that at point ( a, f(a) ) the graph of the function f(x) has no holes or gaps or discontinuity. That is, its graph is unbroken at point (a, f(a))

Here say ,  = 2

And f(a) = 2

Then,  = f(a)

Hence f(x) is continuous at x=a.

Qualitatively, the graph of a function is said to be continuous at x=a if while travelling along the graph of the function and in crossing over the point at x=a either from left to right or from right to left one does not have to lift his pen

Incase one has to lift his pen, then the graph of the function is said to have a discontinuity or break at x=a.

CONTINUITY OF A FUNCTION OF ONE VARIABLE

Function y = f(x) is said to be continuous at point x=a, where a ? domain of f(x)

If f(x) is not continuous at x=a, we say that f(x) is discontinuous at x=a, f(x) will be discontinuous at x=a if any of the following three conditions are satisfied:

i.) f(a) is not defined,

ii.)   and  exists but are not equal

iii.)   and  exists and are equal but are not equal to f(a).

  1. iv) At least one of the limits does not exist.

PROBLEM: Determine if the following function is continuous at x=0 .

F(x) =
SOLUTION:

here F(x) =

RHL

At x=0, let x = 0 + h,

ie,   =  =  = 3.

LHL

At x = 0, let x = 0 – h,

ie,   =  =  = 3.

But f (0) = 0

Therefore, ,   =  = 0.

Thus, f(x) is discontinuous at x.

REMOVING CONTINUITY

There are two types of discontinuities – type-1: removal discontinuity and type-2: non-removal discontinuity.

REMOVABLE DISCONTINUITY

Here, necessarily exists but is either not equal to f(a) or f(a) is not defined.

In this case, therefore, it is possible to redefine the function in such a manner that  and thus making the function continuous.

NON-REMOVAL DISCONTINUITY

Here,  doesn’t exist and therefore, it is not possible to redefine the function in any manner and make it continuous.