Evaluate the following limits, if they exist, where βπ₯β is the greatest integer function.
(a)lim β2π₯β/π₯
π₯β0
(b) lim π₯ β1/π₯β
a:xβ(β12,0):[2x]x=β1xlimβ‘xβ0β[2x]x=ββxβ(0,12):[2x]x=0limβ‘xβ0β[2x]x=0Theββlimβ‘itββdoesnΒtββexistb:xβ (1xβ1)<x[1x]β©½xβ 1xlimβ‘xβ0x(1xβ1)=1,limβ‘xβ0xβ 1x=1ββlimβ‘xβ0x[1x]=1a:\\x\in \left( -\frac{1}{2},0 \right) :\frac{\left[ 2x \right]}{x}=\frac{-1}{x}\\\underset{x\rightarrow 0-}{\lim}\frac{\left[ 2x \right]}{x}=-\infty \\x\in \left( 0,\frac{1}{2} \right) :\frac{\left[ 2x \right]}{x}=0\\\underset{x\rightarrow 0-}{\lim}\frac{\left[ 2x \right]}{x}=0\\The\,\,\lim it\,\,doesnΒt\,\,exist\\b:\\x\cdot \left( \frac{1}{x}-1 \right) <x\left[ \frac{1}{x} \right] \leqslant x\cdot \frac{1}{x}\\\underset{x\rightarrow 0}{\lim}x\left( \frac{1}{x}-1 \right) =1,\underset{x\rightarrow 0}{\lim}x\cdot \frac{1}{x}=1\Rightarrow \\\Rightarrow \underset{x\rightarrow 0}{\lim}x\left[ \frac{1}{x} \right] =1a:xβ(β21β,0):x[2x]β=xβ1βxβ0βlimβx[2x]β=ββxβ(0,21β):x[2x]β=0xβ0βlimβx[2x]β=0ThelimitdoesnΒtexistb:xβ (x1ββ1)<x[x1β]β©½xβ x1βxβ0limβx(x1ββ1)=1,xβ0limβxβ x1β=1ββxβ0limβx[x1β]=1
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