Prove that the greatest integer function `[x]` is continuous at all points except at integer points.

Options

A. N
B. Z
C. R
D. `phi`

Answer

B

Explanation

Let f(x)=[x] be the greatest integer function. Let k be any integer. Then, brgt `f(x)=[x]={{:(,k-1,if k-1 le x le k),(,k,if k le x lt k+1):}," "["By def."]`
Now,
`("LHL at x =k")=underset(x to k^(-))lim f(x)=underset(x to k^(-))lim f(k-h)=underset(x to k^(-))lim[k-h]`
`Rightarrow ("LHL at x=k")=underset(h to 0)lim (k-1)=(k-1)`
and , (RHL at x=k)= `underset(x to k^(+))limf(x)=underset(h to 0)lim f(k+h)=underset(h to 0)lim[k+h]`
`Rightarrow ("RHL at x=k")=underset(h to 0)lim k [therefore k le k + h lt k +1 therefore (k+h)=k]`
`Rightarrow ("RHL at x=k")=k`
`therefore underset(x to k^(-)) f(x)ne underset(x to k^(+))f(x)`
So, f(x) is not continous at x=k
Since k is an arbitrary integer. Therefore, f(x) is a not continous at integer points.
Let a be any real number other than an integer. Then, there exists an integer k such that `k-1 lt a lt k` .
`("LHL at x=a")=underset(x to a^(-))limf(x)=underset(h to 0)limf(a-h)=underset(h to 0)lim[a-h]`
`Rightarrow ("RHL at x=a")=underset(h to 0)lim k-1" "[{:(,therefore,k-1 lt a lt h lt k),(,therefore,[a+h]=k-1):}] `
`Rightarrow ("RHL at x=a")=(k-1) and f(a)=k-1 " "[therefore k-1lt k lt a therefore [a]=k-1]`
Thus, `underset(x-a^(-))lim f(x)=underset(x to a^(+))lim f(x)=f(a)`
So, f(x) is a continous at x=a. Since a is an arbitarary real number, other than an integer, therefore f(x) is a continous at all real points except integer points.