Answer to Question #315584 in Calculus for Aune

Question #315584

Evaluate the following limits, if they exist, where ⌊π‘₯βŒ‹ is the greatest integer function.



(a)lim ⌊2π‘₯βŒ‹/π‘₯


π‘₯β†’0



(b) lim π‘₯ ⌊1/π‘₯βŒ‹


π‘₯β†’0

1
Expert's answer
2022-03-25T05:17:26-0400

a:x∈(βˆ’12,0):[2x]x=βˆ’1xlim⁑xβ†’0βˆ’[2x]x=βˆ’βˆžx∈(0,12):[2x]x=0lim⁑xβ†’0βˆ’[2x]x=0The  lim⁑it  doesnΒ‘t  existb:xβ‹…(1xβˆ’1)<x[1x]β©½xβ‹…1xlim⁑xβ†’0x(1xβˆ’1)=1,lim⁑xβ†’0xβ‹…1x=1β‡’β‡’lim⁑xβ†’0x[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\,\,doesnΒ‘t\,\,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] =1


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