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Question

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{{x}^{3}-3{x}^{2}+2x}\right]$

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Solution

$\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{{x}^{3}-3{x}^{2}+2x}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left({x}^{2}-3x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{-2\left(2x-3\right)}{x\left({x}^{2}-2x-x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left\{x\left(x-2\right)-1\left(x-2\right)\right\}}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{x\left(x-1\right)-2\left(2x-3\right)}{x\left(x-1\right)\left(x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-x-4x+6}{x\left(x-1\right)\left(x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-5x+6}{x\left(x-1\right)\left(x+2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{{x}^{2}-2x-3x+6}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{x\left(x-2\right)-3\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\underset{x\to 2}{\mathrm{lim}}\left[\frac{\left(x-3\right)\left(x-2\right)}{x\left(x-1\right)\left(x-2\right)}\right]\phantom{\rule{0ex}{0ex}}=\frac{2-3}{2\left(2-1\right)}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}$

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