Evaluate \(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)
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Correct Answer: Option A
Explanation:
\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)
\(\frac{1 - x}{x^{2} - 3x + 2} = \frac{-(x - 1)}{(x - 1)(x - 2)}\)
= \(\frac{-1}{x - 2}\)
\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2} = \lim \limits_{x \to 1} \frac{-1}{x - 2}\)
= \(\frac{-1}{1 - 2} = \frac{-1}{-1} = 1\)
\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2}\)
\(\frac{1 - x}{x^{2} - 3x + 2} = \frac{-(x - 1)}{(x - 1)(x - 2)}\)
= \(\frac{-1}{x - 2}\)
\(\lim \limits_{x \to 1} \frac{1 - x}{x^{2} - 3x + 2} = \lim \limits_{x \to 1} \frac{-1}{x - 2}\)
= \(\frac{-1}{1 - 2} = \frac{-1}{-1} = 1\)