Question #87458

For each x ∈ R, let us denote by C(x) the least integer greater than or equal to x.
For example, C(1) = 1, C(−√2) = −1. In other words, C(x) is the unique integer
satisfying C(x) − 1 < x ≤ C(x).
(1) Draw the graph of the function C(x) for x ∈ [−2, 2].
(2) Prove that C(x) is continuous at all non-integer points of R.
(3) Prove that C(x) is discontinuous at all integer points of R.

Expert's answer

Let x be a real number. The ceiling function of x, denoted by ⌈x⌉\lceil x\rceil, is the smallest integer that is larger than or equal to x.

For x∈[−2,2]x\in[-2, 2]


x=−2,f(x)=⌈x⌉=−2x=-2, f(x)=\lceil x\rceil=-2

−2<x≤−1,f(x)=⌈x⌉=−1-2\lt x \le-1, f(x)=\lceil x\rceil=-1

−1<x≤0,f(x)=⌈x⌉=0-1\lt x \le0, f(x)=\lceil x\rceil=0

0<x≤1,f(x)=⌈x⌉=10\lt x \le1, f(x)=\lceil x\rceil=1

1<x≤2,f(x)=⌈x⌉=21\lt x \le2, f(x)=\lceil x\rceil=2




For ai∈(i,i+1),i∈Za_i\in(i, i+1), i\in \Z


lim⁡x→aif(x)=i+1,f(a)=i+1,x∈(i,i+1)\lim\limits_{x\to a_i}f(x)=i+1, f(a)=i+1, x\in(i, i+1)

Therefore, the function f(x)f(x) is continuous at x∈R,x∉Zx\in \R, x\notin \Z



For ai=i∈Z:a_i= i\in \Z:

for x∈(i−1,i)x\in(i-1, i)


lim⁡x→aif(x)=i,\lim\limits_{x\to a_i}f(x)=i,

for x∈(i,i+1)x\in(i, i+1)


lim⁡x→aif(x)=i+1.\lim\limits_{x\to a_i}f(x)=i+1.

Then


lim⁡x→aif(x)=does not exist\lim\limits_{x\to a_i}f(x)=does \ not \ exist

Therefore, the function f(x)f(x) is discontinuous at x∈Z.x\in \Z.


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