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Q. $\displaystyle \lim_{x \to 2}$$\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x^{3}-3x^{2}+2x}\right]$ is equal to

Limits and Derivatives

Solution:

We have, $\displaystyle \lim_{x \to 2}$$\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x^{3}-3x^{2}+2x}\right]$
$=\displaystyle \lim_{x \to 2}$$\left[\frac{1}{x-2}-\frac{2\left(2x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right]$
$=\displaystyle \lim_{x \to 2}$$\left[\frac{x\left(x-1\right)-2\left(2x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right]$
$=\displaystyle \lim_{x \to 2}$$\left[\frac{x^{2}-5x+6}{x\left(x-1\right)\left(x-2\right)}\right]$
$=\displaystyle \lim_{x \to 2}$$\left[\frac{\left(x-2\right)\left(x-3\right)}{x\left(x-1\right)\left(x-2\right)}\right]$
$=\displaystyle \lim_{x \to 2}$$\left[\frac{\left(x-3\right)}{x\left(x-1\right)}\right]=\frac{-1}{2}$