Tìm các giới hạn sau: a) lim (căn 3 /2)^n

Giải SBT Toán 11 Bài 1: Giới hạn của dãy số

Bài 3 trang 75 SBT Toán 11 Tập 1: Tìm các giới hạn sau:

a) $\text{lim}{\left(\frac{\sqrt{3}}{2}\right)}^n;$

b) $\text{lim}\frac{3^n}{4^n-1};$

c) $\text{lim}\frac{3^n-2^n}{3^n+2^n};$

d) $\text{lim}\frac{4^{n+1}}{3^n+4^n}.$

Lời giải:

a) $\text{lim}{\left(\frac{\sqrt{3}}{2}\right)}^n=0.$

b) $\text{lim}\frac{3^n}{4^n-1}=\text{lim}\frac{{\left(\frac{3}{4}\right)}^n}{1-{\left(\frac{1}{4}\right)}^n}=\frac{\text{lim}{\left(\frac{3}{4}\right)}^n}{1-\text{lim}{\left(\frac{1}{4}\right)}^n}=\frac{0}{1-0}=0.$

c) $\text{lim}\frac{3^n-2^n}{3^n+2^n}=\text{lim}\frac{1-{\left(\frac{2}{3}\right)}^n}{1+{\left(\frac{2}{3}\right)}^n}=\frac{1-\text{lim}{\left(\frac{2}{3}\right)}^n}{1+\text{lim}{\left(\frac{2}{3}\right)}^n}=\frac{1-0}{1+0}=1.$

d) $\text{lim}\frac{4^{n+1}}{3^n+4^n}=\text{lim}\frac{4}{{\left(\frac{3}{4}\right)}^n+1}=\frac{4}{\text{lim}{\left(\frac{3}{4}\right)}^n+1}=\frac{4}{0+1}=4$.