Q54 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics

$$ \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}= $$

  • A
    $\frac{1}{5}$
  • B
    $\frac{1}{10}$
  • Correct answer
    C
    $\frac{-1}{10}$
  • D
    $\frac{-1}{5}$

Solution

$$ \begin{aligned} & \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} \\ & \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(x-1)(2 x+3)}=\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)(2 x+3)} \\ & \lim _{x \rightarrow 1} \frac{(2 x-3)}{(\sqrt{x}+1)(2 x+3)}=\frac{-1}{2(5)}=\frac{-1}{10} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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