## Calculus 10th Edition

$\lim\limits_{x\to1}(1-[[-\dfrac{x}{2}]])=2.$
$\lim\limits_{x\to1^+}(1-[[-\dfrac{x}{2}]])=1-[[-\dfrac{1^+}{2}]]=1-(-1)=2.$ $\lim\limits_{x\to1^-}(1-[[-\dfrac{x}{2}]])=1-[[-\dfrac{1^-}{2}]]=1-(-1)=2.$ Since $\lim\limits_{x\to1^+}(1-[[-\dfrac{x}{2}]])=\lim\limits_{x\to1^-}(1-[[-\dfrac{x}{2}]])\to$ $\lim\limits_{x\to1}(1-[[-\dfrac{x}{2}]])=2.$