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Question
\[\lim_{x \to 1} \left\{ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right\}\]
Solution
\[\lim_{x \to 1} \left[ \frac{x - 2}{x^2 - x} - \frac{1}{x^3 - 3 x^2 + 2x} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{x - 2}{x\left( x - 1 \right)} - \frac{1}{x\left( x^2 - 3x + 2 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{x - 2}{x\left( x - 1 \right)} - \frac{1}{x\left( x^2 - 2x - x + 2 \right)} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{x - 2}{x\left( x - 1 \right)} - \frac{1}{x\left\{ x\left( x - 2 \right) - 1\left( x - 2 \right) \right\}} \right]\]
\[ = \lim_{x \to 1} \left[ \frac{\left( x - 2 \right)^2 - 1}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[= \lim_{x \to 1} \left[ \frac{\left( x - 1 \right)\left( x - 3 \right)}{x\left( x - 1 \right)\left( x - 2 \right)} \right]\]
\[ = \frac{\left( 1 - 3 \right)}{1\left( 1 - 2 \right)}\]
\[ = \frac{- 2}{- 1}\]
\[ = 2\]
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