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Q. Every polynomial function is ...$A$... and greatest integer function defined by $f(x)=[x]$ is ...$B$ ... at every integral point. Here, $A$ and $B$ refer to

Continuity and Differentiability

Solution:

Recall that a function $p(x)$ is a polynomial function if it is defined by $p(x)=a_0+a_1 x+\ldots+a_n x^n$ for some natural number $n, a_n \neq 0$ and $a_i \in R$. Clearly, this function is defined for every real number. For a fixed real number $c$, we have
$\displaystyle\lim _{x \rightarrow c} p(x)=p(c)$
By definition, $p$ is continuous at $c$. Since, $c$ is any real number, $p$ is continuous at every real number and hence $p$ is a continuous function.
For $f(x)=[x]$, first observe that $f$ is defined for all real numbers. Graph of the function is given in figure. From the graph it looks like that $f$ is discontinuous at every integral point. Below we explore, if this is true.
image
Case I Let $c$ be a real number which is not equal to any integer. It is evident from the graph that for all real numbers close to $c$ the value of the function is equal to [c] i.e., $\displaystyle\lim _{x \rightarrow c} f(x)=\displaystyle\lim _{x \rightarrow c}[x]=[c]$ and hence the function is continuous at all real numbers not equal to integers.
Case II Let $c$ be an integer. Then, we can find a sufficiently small real number $r>0$ such that $[c-r]=c-1$ whereas $[c+r]=c$.
This, in terms of limits, we have
$\displaystyle\lim _{x \rightarrow c^{-}} f(x)=c-1, \lim _{x \rightarrow c^{+}} f(x)=c$
Since, these limits cannot be equal to each other for any $c$, the function is discontinuous at every integral point.