If `f(x)={(2^(x+2)-16)/(4^x-16),\ \ \ if\ x!=2k ,\ \ \ if\ x=2` is continuous at `x=2` , find `k` .

Explanation

`f(x)=frac{2^{x+2}-16}{4^{z}-16}, text { if } x neq 2 k, text { if } x=2)` If `f(x)` is continuous at `x=2`, then



`lim _{x -> 2} f(x)=f(2)`


`Rightarrow lim _{x -> 2} frac{2^{x+2}-16}{4^{x}-16}=f(2)`


`Rightarrow lim _{x -> 2} frac{4(2^{x}-4)}{(2^{x}-4)(2^{x}+4)}=k`


`Rightarrow lim _{x -> 2} frac{4}{(2^{x}+4)}=k`


`Rightarrow frac{4}{(2^{2}+4)}=k`


`Rightarrow frac{4}{8}=k`


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