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Exercise 5.2 · Q10

Q.Prove that the greatest integer function defined by f(x)=[x]f(x) = [x], 0<x<30 < x < 3, is not differentiable at x=1x = 1 and x=2x = 2.

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[x][x] has a jump at each integer, so it is discontinuous — and therefore not differentiable — at x=1x=1 and x=2x=2; the one-sided derivatives there also disagree.

On 0<x<30<x<3 the greatest integer function is a staircase: [x]=0[x]=0 on (0,1)(0,1), [x]=1[x]=1 on [1,2)[1,2), [x]=2[x]=2 on [2,3)[2,3). Differentiability requires continuity first, and where the graph jumps it cannot be continuous.

Discontinuity forces non-differentiability at x=1x=1

Left: for xx just below 11, [x]=0[x]=0, so lim⁡x→1−[x]=0\lim_{x\to 1^-}[x]=0.

Right: for xx just above 11, [x]=1[x]=1, so lim⁡x→1+[x]=1\lim_{x\to 1^+}[x]=1.

The one-sided limits differ, so lim⁡x→1[x]\lim_{x\to 1}[x] does not exist — ff is discontinuous, hence not differentiable at x=1x=1.

Confirm with the derivative definition at x=1x=1

Using f(1)=[1]=1f(1)=[1]=1:

f+′(1)=lim⁡h→0+[1+h]−1h=lim⁡h→0+1−1h=0,f'_+(1)=\lim_{h\to 0^+}\frac{[1+h]-1}{h}=\lim_{h\to 0^+}\frac{1-1}{h}=0,

f−′(1)=lim⁡h→0−[1+h]−1h=lim⁡h→0−0−1h=lim⁡h→0−−1h→+∞.f'_-(1)=\lim_{h\to 0^-}\frac{[1+h]-1}{h}=\lim_{h\to 0^-}\frac{0-1}{h}=\lim_{h\to 0^-}\frac{-1}{h}\to+\infty.

The right-hand derivative is 00 and the left-hand derivative diverges, so they are unequal — f′(1)f'(1) does not exist. …

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