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Question Number 51095 by Tinkutara last updated on 23/Dec/18

Commented by prakash jain last updated on 25/Dec/18

$$\underset{x\to 0^{+}}{\lim }{\left[x\right]}^2-\left[x^2\right]=0$$$$\underset{x\to 0^{-}}{\lim }{\left[x\right]}^2-\left[x^2\right]$$$$=\underset{h\to 0^{+}}{\lim }{[-h]}^2-\left[{(-h)}^2\right]$$$$=\underset{h\to 0^{+}}{\lim }{[-h]}^2-\left[h^2\right]=-1-0=-1$$$$\mathrm{LHL}\ne \mathrm{RHL}$$

Commented by Tinkutara last updated on 25/Dec/18

And at 1?

Commented by prakash jain last updated on 25/Dec/18

$$\mathrm{You}\mathrm{can}\mathrm{use}\mathrm{the}\mathrm{same}\mathrm{logic}\mathrm{at}1.$$

Commented by prakash jain last updated on 25/Dec/18

$$\underset{x\to 1^{+}}{\lim }{\left[x\right]}^2-\left[x^2\right]=1^2-1^2=0$$$$\underset{x\to 1^{-}}{\lim }{\left[x\right]}^2-\left[x^2\right]=$$$$\underset{h\to 0}{\lim }{\left[(1-h)\right]}^2-\left[{(1-h)}^2\right]$$$$={\left(0\right)}^2-\left(0\right)=0[{(1-h)}^2<1]$$$$\mathrm{LHL}=\mathrm{RHL}=valueoffunction$$

Commented by Tinkutara last updated on 25/Dec/18

Thank you very much Sir! I got the answer. ��������

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