Cho hàm số \(f\left( x \right) = \frac{{\sqrt {x + 1} - 2}}{{x - 3}}\). Khi đó:

a) \(f\left( 8 \right) = - \frac{1}{5}\).

b) \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \frac{1}{3}\).

c) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \frac{1}{6}\).

d) Biết \(\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = a,\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 2} - x} \right) = b\). Khi đó 3a + 4b = 2.

a) \(f\left( 8 \right) = \frac{{\sqrt {8 + 1}  - 2}}{{8 - 3}} = \frac{1}{5}\).

b) \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1}  - 2}}{{x - 3}} = \frac{1}{3}\).

c) \(\mathop {\lim }\limits_{x \to 3} f\left( x \right) = \mathop {\lim }\limits_{x \to 3} \frac{{x - 3}}{{\left( {\sqrt {x + 1}  + 2} \right)\left( {x - 3} \right)}} = \mathop {\lim }\limits_{x \to 3} \frac{1}{{\sqrt {x + 1}  + 2}} = \frac{1}{4}\).

d) \(\mathop {\lim }\limits_{x \to  + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{x\sqrt {\frac{1}{x} + \frac{1}{{{x^2}}}}  - 2}}{{x - 3}} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\sqrt {\frac{1}{x} + \frac{1}{{{x^2}}}}  - \frac{2}{x}}}{{1 - \frac{3}{x}}} = 0\).

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

Do đó 3a + 4b = 2.

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