\(\lim\limits\frac{1+2^n}{2^{n+1}-16}=\lim\limits\frac{\left(\frac{1}{2}\right)^n+1}{2-16\left(\frac{1}{2}\right)^n}=\frac{0+1}{2-0}=\frac{1}{2}\)
\(\lim\limits\left(u_n\right)=\lim\limits\frac{\sqrt{16n^2-n+1}}{3n-2}=\lim\limits\frac{\sqrt{16-\frac{1}{n}+\frac{1}{n^2}}}{3-\frac{2}{n}}=\frac{\sqrt{16-0+0}}{3-0}=\frac{4}{3}\)