TS EAMCET · Maths · Limits
\[ \lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}= \]
- A 0
- B \(\infty\)
- C \(\frac{5}{7}\)
- D \(\frac{7}{9}\)
Answer & Solution
Correct Answer
(D) \(\frac{7}{9}\)
Step-by-step Solution
Detailed explanation
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 :-…
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