Q105 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

$$ \lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}= $$

  • A
    0
  • B
    $\infty$
  • C
    $\frac{5}{7}$
  • Correct answer
    D
    $\frac{7}{9}$

Solution

Given $\lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}$ as $\mathrm{x} \rightarrow 2, \frac{\mathrm{N}^{\mathrm{r}} \rightarrow 0}{\mathrm{D}^{\mathrm{r}} \rightarrow 0} \Rightarrow\left(\frac{0^{\prime}}{0^{\prime}}\right)$ form using L.H. rule :- $\lim _{x \rightarrow 2} \frac{3 x^2-2 x-1}{6 x^2-6 x-3}$ $\Rightarrow \frac{7}{9}$

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

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