Question

Tính: lim\(\left(2n-\sqrt{9n^2+n}+\sqrt{n^2+2n}\right)\)

Answer 1

\(\lim\limits\left(2n-\sqrt{9n^2+n}+\sqrt{n^2+2n}\right)\)

\(=\lim\limits\left(3n-\sqrt{9n^2+n}-n+\sqrt{n^2+2n}\right)\)

\(=\lim\limits\left(\dfrac{9n^2-9n^2-n}{3n+\sqrt{9n^2+n}}-\left(n-\sqrt{n^2+2n}\right)\right)\)

\(=\lim\limits\left(\dfrac{-n}{3n+n\cdot\sqrt{9+\dfrac{1}{n}}}-\dfrac{n^2-n^2-2n}{n+\sqrt{n^2+2n}}\right)\)

\(=\lim\limits\left(-\dfrac{1}{3+\sqrt{9+\dfrac{1}{n}}}+\dfrac{2n}{n+\sqrt{n^2+2n}}\right)\)

\(=\lim\limits\left(-\dfrac{1}{3+\sqrt{9+\dfrac{1}{n}}}+\dfrac{2}{1+\sqrt{1+\dfrac{2}{n}}}\right)\)

\(=\dfrac{-1}{3+\sqrt{9}}+\dfrac{2}{1+\sqrt{1}}=\dfrac{-1}{6}+1=\dfrac{5}{6}\)