Question

please answer this . please please please​

Answer 1

We have,

$\quad \lim _{x \to \frac{1}{2} } \dfrac{8 {x}^{3} - 1}{6 {x}^{2} - 5x + 1} \\ \\ = \dfrac{8 { \left( \frac{1}{2} \right)}^{3} - 1 }{6 { \left( \frac{1}{2} \right)}^{2} - 5 \left( \frac{1}{2} \right) + 1} \\ \\ = \frac{8 \left( \frac{1}{8} \right) - 1}{6 \left( \frac{1}{4} \right) - \frac{5}{2} + 1} \\ \\ = \frac{ \cancel{8} \left( \frac{1}{ \cancel{8}} \right) - 1}{ {}^{3} \cancel{6} \left( \frac{1}{ \cancel{4} \: _2} \right) - \frac{5}{2} + 1} \\ \\ = \frac{1 - 1}{ \frac{3}{2} - \frac{5}{2} + 1} \\ \\ = \frac{ \cancel{1} - \cancel{1}}{ \frac{3 - 5}{2} + 1} \\ \\ = \frac{0}{ - \frac{ \cancel{2}}{ \cancel{2}} + 1 } \\ \\ = \frac{0}{ - 1 + 1} \\ \\ = \frac{0}{0} = \frac{100 - 100}{100 - 100} \\ \\ = \frac{ {(10)}^{2} - {(10)}^{2} }{10(10 - 10)} \\ \\ = \frac{(10 + 10)(10 - 10)}{10(10 - 10)} = \frac{(10 + 10) \cancel{(10 - 10)}}{10 \cancel{(10 - 10)}} \\ \\ = \frac{20}{10} = \frac{2 \cancel{0}}{1 \cancel{0}} \\ \\ = \bf2 \red{ \leftarrow} \large\green{ \bf answer}$

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