Cho hàm số \(f\left( x \right)\) xác định trên \(\mathbb{R}\) liên tục tại \(x = 1\) và thỏa mãn \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = 5\). Khi đó \(f\left( 1 \right)\) bằng bao nhiêu?

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.

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

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

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

\( = \lim \left( {1 - \frac{1}{{\sqrt {n + 1} }}} \right) = 1\).

Trả lời: 1.