# Find the value of k in the following so that the function $f$ is continuous at the indicated point: $f(x)=\left \{\begin{array}{1 1}\frac{\large 2^{x+2}-16}{4^{\large x}-16}, & if\;x\neq 2\\k, & if\;x=2\end{array}\right.$ at $x=2$

$\begin{array}{1 1} k=\frac{1}{2} \\k= \frac{1}{4} \\ k=2 \\ k=4 \end{array}$

Toolbox:
• A function is said to be continuous at a point $'a'$ if the LHL = RHL.
• (i.e) $\lim\limits_{\large x\to a^+}f(x)=\lim\limits_{\large x\to a^-}f(x)$
Step 1:
Consider $f(x)=\large\frac{2^{\Large x+2}-16}{4^{\Large x}-16}$
$\qquad\qquad\;\;\;\;=\large\frac{2^{\Large x}.2^{\Large 2}-16}{2^{\Large 2x}-16}$
$\qquad\qquad\;\;\;\;=\large\frac{4(2^{\Large x}-4)}{(2^{\Large x})^{\Large 2}-16}$
$\qquad\qquad\;\;\;\;=\large\frac{4(2^{\Large x}-4)}{(2^{\Large x})^{\Large 2}-4^{\Large 2}}$
We know that $(a^2-b^2)=(a+b)(a-b)$
$\qquad\qquad\;\;\;\;=\large\frac{4(2^{\Large x}-4)}{(2^{\Large x}-4)(2^{\Large x}+4)}$
$\qquad\qquad\;\;\;\;=\large\frac{4}{(2^{\Large x}+4)}$
Step 2:
For $f(x)$ to be continuous
$\lim\limits_{\large x\to 2}f(x)=f(2)$
$\qquad\quad\;\;=\lim\limits_{\large x\to 2}\large\frac{4}{2^{\Large x}+4}$
$\qquad\quad\;\;=\large\frac{4}{2^{\Large 2}+4}$
$\qquad\quad\;\;=\large\frac{4}{4+4}$
$\qquad\quad\;\;=\large\frac{4}{8}$
$\qquad\quad\;\;=\large\frac{1}{2}$
Step 3:
Since the function is continuous $f(2)=k$ at $x=2$
Therefore $k=\large\frac{1}{2}$