$\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ is equal to

$\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ is equal to
  1. $\frac{2}{3}$
  2. $\frac{-2}{3}$
  3. $\frac{3}{2}$
  4. $\frac{-3}{2}$

Solution

$\begin{aligned} & \lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x\left(x^2-3 x+2\right)}\right] \\ & =\lim _{x \rightarrow 2}\left[\frac{1}{(x-2)}-\frac{2}{x(x-2)(x-1)}\right] \\ & =\lim _{x \rightarrow 2}\left[\frac{x^2-x-2}{x(x-1)(x-2)}\right] \\ & =\lim _{x \rightarrow 2} \frac{(x-2)(x+1)}{x(x-1)(x-2)} \\ & =\lim _{x \rightarrow 2} \frac{x+1}{x(x-1)} \\ & =\frac{3}{2}\end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 1)

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