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Question
\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\]
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Solution
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x^2 + 2x + 4 \right)}{\left( x - 2 \right)\left[ x + 2 \right]} \right] \binom{ \because A^3 - B^3 = \left( A - B \right)\left( A^2 + AB + B^2 \right)}{ \because A^2 - B^2 = \left( A - B \right)\left( A + B \right)}\]
\[ = \frac{2^2 + 2 \times 2 + 4}{2 + 2}\]
\[ = \frac{12}{4}\]
\[ = 3\]
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