Find the value of k for which the function f defined as:

f(x)={ 2^x+2 -16/4^x -16 , x not equal to 2
          k , x=2
is continuous at x=2.

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}\frac{2^{\mathrm{x}+2}-16}{4^{\mathrm{x}}-16}& \mathrm{x}\ne 2\\ \mathrm{k}& \mathrm{x}=2\end{array}\right. \\ \mathrm{Function} \mathrm{will} \mathrm{be} \mathrm{continuous} \mathrm{at} \mathrm{x}=2 \mathrm{if} \\ \underset{\mathrm{x}\to 2}{\lim } \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(2\right) \\ \Rightarrow \underset{\mathrm{x}\to 2}{\lim } \mathrm{f}\left(\mathrm{x}\right)=\mathrm{k} \\ \mathrm{Now} \\ \underset{\mathrm{x}\to 2}{\lim } \mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 2}{\lim } \frac{2^{\mathrm{x}+2}-16}{4^{\mathrm{x}}-16} \\ =\underset{\mathrm{x}\to 2}{\lim } \frac{2^{\mathrm{x}}.2^2-16}{2^{2\mathrm{x}}-4^2} \\ =\underset{\mathrm{x}\to 2}{\lim } \frac{4.2^{\mathrm{x}}-16}{{\left(2^{\mathrm{x}}\right)}^2-4^2} \\ =4\underset{\mathrm{x}\to 2}{\lim } \frac{2^{\mathrm{x}}-4}{\left(2^{\mathrm{x}}-4\right)\left(2^{\mathrm{x}}+4\right)} \\ =4\underset{\mathrm{x}\to 2}{\lim } \frac{1}{\left(2^{\mathrm{x}}+4\right)} \\ =4 \frac{1}{2^2+4} \\ =\frac{4}{4+4} \\ =\frac{1}{2} \left[\mathrm{Answer}\right] \end{gathered}$$

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