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.

Answer 1

Given:

\[f\left( x \right) = \binom{\frac{2^{x + 2} - 16}{4^x - 16}, \text{ if }  x \neq 2}{k , \text{ if }  x = 2}\] If f(x) is continuous at x = 2, then

\[\lim_{x \to 2} f\left( x \right) = f\left( 2 \right)\]
\[ \Rightarrow \lim_{x \to 2} \frac{2^{x + 2} - 16}{4^x - 16} = f\left( 2 \right)\]
\[ \Rightarrow \lim_{x \to 2} \frac{4\left( 2^x - 4 \right)}{\left( 2^x - 4 \right)\left( 2^x + 4 \right)} = k\]
\[ \Rightarrow \lim_{x \to 2} \frac{4}{\left( 2^x + 4 \right)} = k\]
\[ \Rightarrow \frac{4}{\left( 2^2 + 4 \right)} = k\]
\[ \Rightarrow \frac{4}{8} = k\]
\[ \Rightarrow k = \frac{1}{2}\]