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Question
\[\lim_{x \to 2} \frac{x^4 - 16}{x - 2}\]
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Solution
\[\lim_{x \to 2} \left[ \frac{x^4 - 16}{x - 2} \right]\]
\[\text{ It is of the form } \frac{0}{0} . \]
\[ \lim_{x \to 2} \left[ \frac{\left( x^2 \right)^2 - \left( 4 \right)^2}{x - 2} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{\left( x^2 - 4 \right)\left( x^2 + 4 \right)}{x - 2} \right]\]
\[ = \lim_{x \to 2} \left[ \frac{\left( x - 2 \right)\left( x + 2 \right)\left( x^2 + 4 \right)}{x - 2} \right]\]
\[ = \left( 2 + 2 \right)\left( 2^2 + 4 \right)\]
\[ = 32\]
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