Câu 7:
$\lim\dfrac{7-6\sqrt{n+3}}{\sqrt{4n+5}+\sqrt{n-1}}$
$=\lim\dfrac{\dfrac{7}{\sqrt{n}}-6\sqrt{1+\dfrac{3}{n}}}{\sqrt{4+\dfrac{5}{n}}+\sqrt{1-\dfrac{1}{n}}}$
$=\dfrac{0-6\sqrt{1+0}}{2+1-0}=-2$
Câu 8:
$\lim\dfrac{\sqrt[3]{n^3-8n}+2n-3}{\sqrt{4n^2-1}-n-5}$
$=\lim\dfrac{\sqrt[3]{1-\dfrac{8}{n^2}}+2-\dfrac{3}{n}}{\sqrt{4-\dfrac{1}{n^2}}-1-\dfrac{5}{n}}$
$=\dfrac{1+2-0}{2-1+0}=3$
