MHT CET · Maths · Limits
\(\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]\) is equal to
- A \(\frac{2}{3}\)
- B \(\frac{-2}{3}\)
- C \(\frac{3}{2}\)
- D \(\frac{-3}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x\left(x^2-3 x+2\right)}\right] \\ & =\lim _{x \rightarrow 2}\left[\frac{1}{(x-2)}-\frac{2}{x(x-2)(x-1)}\right] \\ & =\lim _{x \rightarrow 2}\left[\frac{x^2-x-2}{x(x-1)(x-2)}\right] \\ & =\lim _{x \rightarrow 2} \frac{(x-2)(x+1)}{x(x-1)(x-2)} \\ & =\lim _{x \rightarrow 2} \frac{x+1}{x(x-1)} \\ & =\frac{3}{2}\end{aligned}\)
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