Question

Tìm giới hạn dãy số :

\(a,lim\dfrac{5n+1}{2n}\\ b,lim\dfrac{6n^2+8n+1}{5n^2+3}\\ c,lim\dfrac{3^n+2^n}{4.3^n}\\ d,lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}\)

Answer 1

a: \(\lim\limits\dfrac{5n+1}{2n}=\lim\limits\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=\lim\limits\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5+0}{2}=\dfrac{5}{2}\)

b: \(\lim\limits\dfrac{6n^2+8n+1}{5n^2+3}\)

\(=\lim\limits\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}\)

\(=\lim\limits\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}\)

\(=\dfrac{6+0+0}{5+0}=\dfrac{6}{5}\)

c: \(\lim\limits\dfrac{3^n+2^n}{4\cdot3^n}\)

\(=\lim\limits\dfrac{\dfrac{3^n}{3^n}+\left(\dfrac{2}{3}\right)^n}{4\cdot\left(\dfrac{3^n}{3^n}\right)}\)

\(=\lim\limits\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1+0}{4}=\dfrac{1}{4}\)

d: \(\lim\limits\dfrac{\sqrt{n^2+5n+3}}{6n+2}\)

\(=\lim\limits\dfrac{\sqrt{\dfrac{n^2}{n^2}+\dfrac{5n}{n^2}+\dfrac{3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}\)

\(=\lim\limits\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}\)

\(=\dfrac{\sqrt{1+0+0}}{6}=\dfrac{1}{6}\)

Answer 2

\(a,lim\dfrac{5n+1}{2n}=lim\dfrac{\dfrac{5n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}}=lim\dfrac{5+\dfrac{1}{n}}{2}=\dfrac{5}{2}\\ b,lim\dfrac{6n^2+8n+1}{5n^2+3}=lim\dfrac{\dfrac{6n^2}{n^2}+\dfrac{8n}{n^2}+\dfrac{1}{n^2}}{\dfrac{5n^2}{n^2}+\dfrac{3}{n^2}}=lim\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}=\dfrac{6}{5}\)

\(c,lim\dfrac{3^n+2^n}{4.3^n}=\dfrac{\dfrac{3^n}{3^n}+\dfrac{2^n}{3^n}}{\dfrac{4.3^n}{3^n}}=\dfrac{1+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1}{4}\)

\(d,lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}=lim\dfrac{\sqrt{\dfrac{n^2+5n+3}{n^2}}}{\dfrac{6n}{n}+\dfrac{2}{n}}=lim\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}=\dfrac{1}{6}\)

Answer 3

\(a\text{)}lim\dfrac{5n+1}{2n}=lim\dfrac{5}{2}+lim\dfrac{1}{2n}=\dfrac{5}{2}\)

\(b\text{)}lim\dfrac{6n^2+8n+1}{5n^2+3}=lim\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}=\dfrac{6}{5}\)

\(c\text{)}lim\dfrac{3^n+2^n}{4.3^n}=lim\dfrac{\left(\dfrac{3}{3}\right)^n+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1}{4}\)

\(d\text{)}lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}=lim\dfrac{n\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{n\left(6+\dfrac{2}{n}\right)}=lim\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}=\dfrac{1}{6}\)