Question

Prove ceiling function f (x) = ⌈x⌉ is discontinuous for all n ∈ Z.

Answer 1

we have to prove that ceiling function, f(x) = ⌈x⌉ is a discontinuous function for all n ∈ Z.

actually, ceiling function is nothing but least integer function.

⌈x⌉ = n if and only if n - 1 < x ≤ n

for instance, ⌈2.4⌉ = 3 , as 3 - 1 < 2.4 ≤ 3

concept : any function y = f(x) will be continuous at x = a, only if

$\displaystyle\lim_{x\to a^+}f(x)=\displaystyle\lim_{x\to a^-}f(x)=f(a)$

let's take an integer , Z = 1

f(1) = ⌈1⌉ = 1

but, $\displaystyle\lim_{x\to1^+}f(x)=2$ while $\displaystyle\lim_{x\to 1^-}f(x)=1$

i.e., $\displaystyle\lim_{x\to a^+}f(x)\neq\displaystyle\lim_{x\to a^-}f(x)f(a)$

hence, ceiling function is discontinuous at x = 1. similarly you can check all integers. you will get the this function is discontinuous for each integer.

hence, ceiling function f(x) = ⌈x⌉ is a discontinuous function for all n ∈ Z