a,
$\lim\dfrac{2n^2+n}{n^2+4}=\lim\dfrac{2+\dfrac{1}{n}}{1+\dfrac{4}{n^2}}=\dfrac{2+0}{1+0}=2$
b,
$\lim\dfrac{2.3^n+5^n}{3^n+5^n}=\lim\dfrac{2.0,6^n+1}{0,6^n+1}=\dfrac{0+1}{0+1}=1$
c,
$\lim(\sqrt{4n^2+4n}-2n)$
$=\lim\dfrac{4n^2+4n-4n^2}{\sqrt{4n^2+4n}+2n}$
$=\lim\dfrac{4n}{\sqrt{4n^2(1+\dfrac{1}{n})}+2n}$
$=\lim\dfrac{4n}{2n(\sqrt{1+\dfrac{1}{n}}+1)}$
$=\dfrac{2}{2}=1$
