Question:

$\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

Updated On: Jul 29, 2023
  • $-\frac{1}{2}$
  • $\frac{1}{2}$
  • $0$
  • $\infty$
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The Correct Option is A

Solution and Explanation

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}$

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