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Mathematics Question on limits and derivatives

lim⁡x→2\displaystyle \lim_{x \to 2} [1x−2−2(2x−3)x3−3x2+2x]\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x^{3}-3x^{2}+2x}\right] is equal to

A

−12-\frac{1}{2}

B

12\frac{1}{2}

C

00

D

∞\infty

Answer

−12-\frac{1}{2}

Explanation

Solution

We have, lim⁡x→2\displaystyle \lim_{x \to 2} [1x−2−2(2x−3)x3−3x2+2x]\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x^{3}-3x^{2}+2x}\right] =lim⁡x→2=\displaystyle \lim_{x \to 2} [1x−2−2(2x−3)x(x−1)(x−2)]\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right] =lim⁡x→2=\displaystyle \lim_{x \to 2} [x(x−1)−2(2x−3)x(x−1)(x−2)]\left[\frac{x\left(x-1\right)-2\left(2x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right] =lim⁡x→2=\displaystyle \lim_{x \to 2} [x2−5x+6x(x−1)(x−2)]\left[\frac{x^{2}-5x+6}{x\left(x-1\right)\left(x-2\right)}\right] =lim⁡x→2=\displaystyle \lim_{x \to 2} [(x−2)(x−3)x(x−1)(x−2)]\left[\frac{\left(x-2\right)\left(x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right] =lim⁡x→2=\displaystyle \lim_{x \to 2} [(x−3)x(x−1)]=−12\left[\frac{\left(x-3\right)}{x\left(x-1\right)}\right]=\frac{-1}{2}