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  • Let f(x)=[x] and <img alt="\mathrm{g(x)=\left\{\begin{array}{ll}0, & x \in Z \\ x^2, & x \in R-Z\end{array}\right.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bg%

Let f(x)=[x] and \mathrm{g(x)=\left\{\begin{array}{ll}0, & x \in Z \\ x^2, & x \in R-Z\end{array}\right.}. Then

Option: 1

\mathrm{\lim _{x \rightarrow 1} } g(x) exists, but g(x) is not continuous at x=2.


Option: 2

\mathrm{\lim _{x \rightarrow 1}} f(x) does not exist and f(x) is not continuous at x=1.


Option: 3

gof is continuous for all f(x)


Option: 4

fog is continuous for all x.


Answers (1)

best_answer

Since, \mathrm{\lim _{x \rightarrow 1^{-}} g(x)=\lim _{x \rightarrow 1^{+}} g(x)=1} and g(1)=0. So

g(x) is not continuous at x=1 but \mathrm{\lim _{x \rightarrow 1}} g(x) exists

We have, \mathrm{\lim _{\mathrm{x} \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}[1-h]=0}

and, \mathrm{\quad \lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}[1+h]=1}

So, \mathrm{\lim _{x \rightarrow 1} f(x)} does not exist and so f(x) is not continuous at x=1.

We have, \mathrm{g \circ f(x)=g(f(x))=g([x])=0, \forall x, y x \in R}

So, gof is continuous for all x.
We have,
\mathrm{ f \circ g(x)=f(g(x))) = \begin{cases}f(0), & x \in Z \\ f\left(x^2\right), & x \in R-Z\end{cases} }
                                        \mathrm{ = \begin{cases}0, & x \in Z \\ {\left[x^2\right],} & x \in R-Z\end{cases} }
which is clearly not continuous.

Posted by

Divya Prakash Singh

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