Question

Tìm \(\lim\left(n+\sqrt{n^2-n+1}\right)\).

Answer 1

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

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

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

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

\(=\lim\limits\dfrac{1-\dfrac{1}{n}}{1-\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}}=+\infty\)

Vì \(\left\{{}\begin{matrix}\lim\limits1-\dfrac{1}{n}=1-0=1\\\lim\limits\left(1-\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}\right)=1-\sqrt{1-0+0}=1-1=0\end{matrix}\right.\)