Answer
$\lim\limits_{x\to 1}{\frac{1-x^2}{x^2-8x+7}} = 3$
Work Step by Step
$\lim\limits_{x\to 1}{\frac{1-x^2}{x^2-8x+7}} = \lim\limits_{x\to 1}{\frac{(1-x)(1+x)}{(x-7)(x-1)}} =\lim\limits_{x\to 1}{\frac{-(x-1)(1+x)}{(x-7)(x-1)}} = \lim\limits_{x\to 1}{\frac{-(1+x)}{(x-7)}} =\frac{-(1+\lim\limits_{x\to 1}x)}{(\lim\limits_{x\to 1}x-7)} = \frac{-2}{-6}=3$