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Question
If \[f\left( x \right) = \begin{cases}\frac{2^{x + 2} - 16}{4^x - 16}, \text{ if } & x \neq 2 \\ k , \text{ if } & x = 2\end{cases}\] is continuous at x = 2, find k.
Solution
Given:
\[\lim_{x \to 2} f\left( x \right) = f\left( 2 \right)\]
\[ \Rightarrow \lim_{x \to 2} \frac{2^{x + 2} - 16}{4^x - 16} = f\left( 2 \right)\]
\[ \Rightarrow \lim_{x \to 2} \frac{4\left( 2^x - 4 \right)}{\left( 2^x - 4 \right)\left( 2^x + 4 \right)} = k\]
\[ \Rightarrow \lim_{x \to 2} \frac{4}{\left( 2^x + 4 \right)} = k\]
\[ \Rightarrow \frac{4}{\left( 2^2 + 4 \right)} = k\]
\[ \Rightarrow \frac{4}{8} = k\]
\[ \Rightarrow k = \frac{1}{2}\]
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