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Question
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
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Solution
\[\lim_{x \to - 1/2} \left[ \frac{8 x^3 + 1}{2x + 1} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to - 1/2} \left[ \frac{\left( 2x \right)^3 + 1}{2x + 1} \right]\]
\[ = \lim_{x \to - 1/2} \left[ \frac{\left( 2x + 1 \right)\left\{ \left( 2x \right)^2 - 2x \times 1 + 1^2 \right\}}{\left( 2x + 1 \right)} \right] \left[ \because A^3 + B^3 = \left( A + B \right)\left( A^2 - AB + B^2 \right) \right]\]
\[ = \lim_{x \to - 1/2} \left[ \left( 2x \right)^2 - 2x + 1 \right]\]
\[ = \left( 2 \times \frac{- 1}{2} \right)^2 - 2 \times \frac{- 1}{2} + 1\]
\[ = 1 + 1 + 1\]
\[ = 3\]
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