# Show that the function defined by g(x)=x-[x] is discontinuous at all integral points.

Here [x] denotes the greatest integer less than or equal to x.

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Hint-To solve these types of problems calculate the value of LHL and RHL and show

that the value of $LHL \ne RHL$which means to say that they are discontinuous.

The given function is g(x)=x-[x]

In this function let us consider an integer n and solve it

On substituting the value of n in the equation, we get

g(n)=n-[n]=n-n=0

Now let us take the LHL and RHL of this equation,

We get LHL at x=n=$\mathop {\lim }\limits_{x \to {n^ - }} g(x) = \mathop {\lim }\limits_{x \to

{n^ - }} (x - [x]) = n - (n - 1) = 1$

RHL at x=n=$\mathop {\lim }\limits_{x \to {n^ + }} g(x) = \mathop {\lim }\limits_{x \to

{n^ + }} (x - [x]) = n - n = 0$

So, from this we can clearly observe that the value of $LHL \ne RHL$

If, for a function $LHL \ne RHL$, then we can say that the function is discontinuous

So, we can say that g(x)=x-[x] is discontinuous at all integral points

Note: If a similar type of question is asked to show that the functions are continuous then

show that LHL=RHL , which means to say that the function is continuous.

that the value of $LHL \ne RHL$which means to say that they are discontinuous.

The given function is g(x)=x-[x]

In this function let us consider an integer n and solve it

On substituting the value of n in the equation, we get

g(n)=n-[n]=n-n=0

Now let us take the LHL and RHL of this equation,

We get LHL at x=n=$\mathop {\lim }\limits_{x \to {n^ - }} g(x) = \mathop {\lim }\limits_{x \to

{n^ - }} (x - [x]) = n - (n - 1) = 1$

RHL at x=n=$\mathop {\lim }\limits_{x \to {n^ + }} g(x) = \mathop {\lim }\limits_{x \to

{n^ + }} (x - [x]) = n - n = 0$

So, from this we can clearly observe that the value of $LHL \ne RHL$

If, for a function $LHL \ne RHL$, then we can say that the function is discontinuous

So, we can say that g(x)=x-[x] is discontinuous at all integral points

Note: If a similar type of question is asked to show that the functions are continuous then

show that LHL=RHL , which means to say that the function is continuous.

Last updated date: 28th Sep 2023

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