Question

Tìm các giới hạn sau:

\(a,lim\left(\sqrt{4n^2+5n}-2n\right)\)

\(b,lim\left(\sqrt{2n+1}-\sqrt{n}\right)\)

 

Answer 1

\(\lim\left(\sqrt{4n^2+5n}-2n\right)=\lim\dfrac{5n}{\sqrt{4n^2+5n}+2n}=\lim\dfrac{5}{\sqrt{4+\dfrac{5}{n}}+2}=\dfrac{5}{\sqrt{4+0}+2}=\dfrac{5}{4}\)

\(\lim\left(\sqrt{2n+1}-\sqrt{n}\right)=\lim\sqrt{n}\left(\sqrt{2+\dfrac{1}{n}}-1\right)=+\infty.\left(\sqrt{2}-1\right)=+\infty\) (do \(\sqrt{2}-1>0\))

Answer 2

\(a,lim\left(\sqrt{4n^2+5n}-2n\right)\)

\(=limn\left(\sqrt{4+\dfrac{5}{n}}-2\right)=n.0=0\)

\(b,lim\left(\sqrt{2n+1}-\sqrt{n}\right)\)

\(=lim\sqrt{n}\left(\sqrt{2+\dfrac{1}{n}}-1\right)=\sqrt{n}\left(\sqrt{2}-1\right)=+\infty\)