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प्रश्न
\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]
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उत्तर
\[\lim_{x \to - \frac{1}{2}} \left[ \frac{8 x^3 + 1}{2x + 1} \right]\]
\[ = \lim_{x \to - \frac{1}{2}} \left[ \frac{\left( 2x \right)^3 - \left( - 1 \right)}{2x - \left( - 1 \right)} \right]\]
\[\text{ When } x \to - \frac{1}{2}, \text{ then } 2x \to -1 .\]
Let y = 2x
\[\lim_{y \to - 1} \left[ \frac{y^3 - \left( - 1 \right)^3}{y - \left( - 1 \right)} \right]\]
\[ = 3 \left( - 1 \right)^{3 - 1} \]
\[ = 3 \times \left( - 1 \right)^2 \]
\[ = 3\]
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