Answer to Question #318666 in Calculus for Nameya

Question #318666

Evaluate the following limits,if they exist,where [x] is the greatest interger function


a)lim [2x]/x as x approaches 0


b)lim x[1/x] as x approaches 0


1
Expert's answer
2022-03-28T16:04:26-0400

a:x(12,0):[2x]x=1xlimx0[2x]x=x(0,12):[2x]x=0limx0+[2x]x=0The  limit  doesn‘t  existb:x(1x1)<x[1x]x1xlimx0x(1x1)=1,limx0x1x=1limx0x[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\,\,doesn‘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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