Prove that


$f(x)=\left\{\begin{array}{ll}\frac{2^{x+2}-16}{4^{x}-16}, & \text { if } x \neq 2 \\ k, & \text { if } x=2\end{array}\right.$ at $x=2$


The given function f(x) can be rewritten as,

$f(x)=\frac{2^{x+2}-16}{4^{x}-16}=\frac{2^{2} \cdot 2^{x}-16}{\left(2^{x}\right)^{2}-(4)^{2}}=\frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}$


$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{h \rightarrow 0} \frac{4}{2^{2-h}+4}=\frac{4}{2^{2}+4}=\frac{4}{4+4}=\frac{4}{8}=\frac{1}{2}$

$\lim _{x \rightarrow 2} f(x)=k$

As the function is continuous at $x=2$.

$\therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2} f(x)$

So, k = ½

Therefore, the value of k is ½

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