Bài 5.21 trang 123 SGK Toán 11 tập 1 - Kết nối tri thức

Ta có:

\(\mathop {\lim }\limits_{x \to {\rm{\;}} + \infty } \left( {\sqrt {x + 1} {\rm{\;}} - \sqrt {x + 2} } \right) = \mathop {\lim }\limits_{x \to {\rm{\;}} + \infty } \frac{{\left( {\sqrt {x + 1} {\rm{\;}} - \sqrt {x + 2} } \right)\left( {\sqrt {x + 1}  + \sqrt {x + 2} } \right)}}{{\sqrt {x + 1} {\rm{\;}} + \sqrt {x + 2} }}\)

\( = \mathop {\lim }\limits_{x \to {\rm{\;}} + \infty } \frac{{x + 1 - x - 2}}{{\sqrt {x + 1} {\rm{\;}} + \sqrt {x + 2} }} = \mathop {\lim }\limits_{x \to {\rm{\;}} + \infty } \frac{{ - 1}}{{\sqrt {x + 1} {\rm{\;}} + \sqrt {x + 2} }}\)

\( = \mathop {\lim }\limits_{x \to {\rm{\;}} + \infty } \frac{{ - 1}}{{x\sqrt {1 + \frac{1}{x}} {\rm{\;}} + x\sqrt {1 + \frac{2}{x}} }} = \mathop {\lim }\limits_{x \to {\rm{\;}} + \infty } \frac{{\frac{{ - 1}}{x}}}{{\sqrt {1 + \frac{1}{x}} {\rm{\;}} + \sqrt {1 + \frac{2}{x}} }} = \frac{0}{2} = 0\).

Đáp án: B