Toán 11 Tính giới hạn

a) [tex]lim(\sqrt{n^2+3n}-n+1)=lim(\frac{3n}{\sqrt{n^2+3n}+n}+1)=lim(\frac{3}{\sqrt{1+\frac{3}{n}}+1}+1)=\frac{lim3}{lim(\sqrt{1+\frac{3}{n}}+1)}+lim1=\frac{3}{1+1}=\frac{3}{2}[/tex]
b) [tex]\lim(\sqrt{n+1}-\sqrt{n}).n=\lim \frac{n}{\sqrt{n+1}+\sqrt{n}}=\lim\frac{\sqrt{n}}{\sqrt{1+\frac{1}{n}}+1}=\frac{lim\sqrt{n}}{lim(\sqrt{1+\frac{1}{n}}+1)}=\frac{lim\sqrt{n}}{2}=+\infty[/tex]
c) [tex]\lim{\frac{1}{\sqrt{n^2+2}-\sqrt{n^2-1}}}=\lim \frac{\sqrt{n^2+2}+\sqrt{n^2-1}}{3}=\frac{lim(\sqrt{n^2+2}+\sqrt{n^2-1})}{3}=+\infty[/tex]
d) x tiến dần tới [TEX]-\infty[/TEX] hay [TEX]+\infty[/TEX] nhỉ?