Cho hàm số \(f\left( x \right) = \frac{{\sqrt {3{x^2} + 1} }}{{x + 1}}\) và \(g\left( x \right) = \frac{{2x}}{{x + 1}}\). Xét tính đúng sai của các khẳng định sau:

Hướng dẫn giải:

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

a) Sai. Ta có: \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {3{x^2} + 1} }}{{x + 1}} = \frac{{\sqrt {3 \cdot {1^2} + 1} }}{{1 + 1}} = 1\)

\(\mathop {\lim }\limits_{x \to - 2} g\left( x \right) = \mathop {\lim }\limits_{x \to - 2} \frac{{2x}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 2} \frac{{2 \cdot \left( { - 2} \right)}}{{ - 2 + 1}} = 4\).

b) Đúng. Ta có: \(\mathop {\lim }\limits_{x \to + \infty } g\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \frac{{2x}}{{x + 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{2}{{1 + \frac{1}{x}}} = 2\).

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

c) Sai. Ta có: \(\mathop {\lim }\limits_{x \to - \infty } \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) - \mathop {\lim }\limits_{x \to - \infty } g\left( x \right) = - \sqrt 3 - 2\).

d) Đúng. Ta có:

\(\mathop {\lim }\limits_{x \to - 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to - 1} \frac{{\sqrt {3{x^2} + 1} + 2x}}{{x + 1}} = \mathop {\lim }\limits_{x \to - 1} \frac{{\left( {\sqrt {3{x^2} + 1} + 2x} \right) \cdot \left( {\sqrt {3{x^2} + 1} - 2x} \right)}}{{\left( {x + 1} \right)\left( {\sqrt {3{x^2} + 1} - 2x} \right)}}\)

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

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