Tính \(\lim \left( {\frac{1}{{2\sqrt 1  + 1\sqrt 2 }} + \frac{1}{{3\sqrt 2  + 2\sqrt 3 }} + ... + \frac{1}{{\left( {n + 1} \right)\sqrt n  + n\sqrt {n + 1} }}} \right)\)

Hàm số \(f\left( x \right)\) xác định trên \(\mathbb{R}\) liên tục tại \(x = 1\) \( \Leftrightarrow \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) = 5\). Chọn D.

a) Khi \(m =  - 1\) thì \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 1 - 2} \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 3} \right) = 1\).

b) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \sqrt {x + 7}  = \sqrt {10} \).

c) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x + 7}  = 3\); \(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ - }} \left( {{x^2} - 1 + 2m} \right) = 3 + 2m\); \(f\left( 2 \right) = 3\).

Để tồn tại \(\mathop {\lim }\limits_{x \to 2} f\left( x \right)\) thì \(3 + 2m = 3 \Leftrightarrow m = 0\).

d) \(\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} \sqrt {x + 7}  = 3\).

Đáp án: a) Đúng;    b) Sai;   c) Sai;   d) Đúng.