Given: lim\(_{x→1}\) \(\frac{1-x}{x^2 - 3x + 2}\)
x\(^2\) - 3x + 2
x\(^2\) - 3x + 2 = x\(^2\) - 2x - x + 2
x\(^2\) - 2x - x + 2 = x(x - 2) - 1(x - 2)
= (x - 1)(x -2)
lim\(_{x→1}\) \(\frac{1-x}{x^2 - 3x + 2}\) = \(\frac{1-x}{(x - 1)(x -2)}\)
= \(\frac{-1}{x -2}\)
x → 1
\(\frac{-1}{x -2}\) = \(\frac{-1}{1 -2}\) = \(\frac{-1}{-1}\) = 1
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