Tìm giới hạn I=lim từ x đến dương vô cùng (x+1-căn bậc hai(x^2-x+2)) A.I=1/2 B.I=46/31

Ta có: $I=\underset{x\to +\infty }{\lim }\left(x+1-\sqrt{x^2-x+2}\right)\iff I=\underset{x\to +\infty }{\lim }\left(\frac{x^2-x^2+x-2}{x+\sqrt{x^2-x+2}}+1\right)$

$\iff I=\underset{x\to +\infty }{\lim }\left(\frac{x-2}{x+\sqrt{x^2-x+2}}+1\right)$

$\iff I=\underset{x\to +\infty }{\lim }\left(\frac{1-\frac{2}{x}}{1+\sqrt{1-\frac{1}{x}+\frac{2}{x^2}}}+1\right)\iff I=\frac{3}{2}$