Question

tìm giới hạn sau : lim\(\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}\)

mong các bn và các thầy cô giúp với ạ

Answer 1

\(=lim\dfrac{\left(\sqrt{3n+2}+\sqrt{2n+1}\right)}{3n+2-\left(2n+1\right)}\)

\(=lim\dfrac{\left(\sqrt{3n+2}+\sqrt{2n+1}\right)}{n+1}\)

\(=lim\dfrac{n[\left(\sqrt{\dfrac{3}{n}+\dfrac{2}{n^2}}\right)+\left(\sqrt{\dfrac{2}{n}+\dfrac{1}{n^2}}\right)]}{n\left(1+\dfrac{1}{n^2}\right)}\)

*\(lim[\left(\sqrt{\dfrac{3}{n}+\dfrac{2}{n^2}}\right)+\left(\sqrt{\dfrac{2}{n}+\dfrac{1}{n^2}}\right)]=0\)

*\(lim\left(1+\dfrac{1}{n^2}\right)=1\)

Vậy: \(lim\dfrac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=0\)