Answer
$\lim\limits_{x\to1^-}\dfrac{x^2+2x+1}{x-1}=-\infty.$
Work Step by Step
$\lim\limits_{x\to1^-}\dfrac{x^2+2x+1}{x-1}=\dfrac{(1^-)^2+2(1^-)+1}{(1^-)-1}=\dfrac{4}{0^-}=-\infty.$
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