[LỜI GIẢI] Cho Hàm Số F(x) = Căn X^2 + 2x + 4 - Tự Học 365

Lời giải của Tự Học 365

Giải chi tiết:

\(f(x) = \sqrt {{x^2} + 2x + 4}  - \sqrt {{x^2} - 2x + 4} \)

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  + \infty } f(x) = \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt {{x^2} + 2x + 4}  - \sqrt {{x^2} - 2x + 4} } \right)\\= \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {\sqrt {{x^2} + 2x + 4}  - \sqrt {{x^2} - 2x + 4} } \right)\left( {\sqrt {{x^2} + 2x + 4}  + \sqrt {{x^2} - 2x + 4} } \right)}}{{\left( {\sqrt {{x^2} + 2x + 4}  + \sqrt {{x^2} - 2x + 4} } \right)}}\\ = \mathop {\lim }\limits_{x \to  + \infty } \frac{{({x^2} + 2x + 4) - ({x^2} - 2x + 4)}}{{\sqrt {{x^2} + 2x + 4}  + \sqrt {{x^2} - 2x + 4} }} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{4x}}{{\sqrt {{x^2} + 2x + 4}  + \sqrt {{x^2} - 2x + 4} }}\\= \mathop {\lim }\limits_{x \to  + \infty } \frac{4}{{\sqrt {1 + \frac{2}{x} + \frac{4}{{{x^2}}}}  + \sqrt {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} }} = 2\end{array}\)

 

\(\begin{array}{l}\mathop {\lim }\limits_{x \to  - \infty } f(x) = \mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt {{x^2} + 2x + 4}  - \sqrt {{x^2} - 2x + 4} } \right)\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\left( {\sqrt {{x^2} + 2x + 4}  - \sqrt {{x^2} - 2x + 4} } \right)\left( {\sqrt {{x^2} + 2x + 4}  + \sqrt {{x^2} - 2x + 4} } \right)}}{{\left( {\sqrt {{x^2} + 2x + 4}  + \sqrt {{x^2} - 2x + 4} } \right)}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{({x^2} + 2x + 4) - ({x^2} - 2x + 4)}}{{\sqrt {{x^2} + 2x + 4}  + \sqrt {{x^2} - 2x + 4} }} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{4x}}{{\sqrt {{x^2} + 2x + 4}  + \sqrt {{x^2} - 2x + 4} }}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{{\frac{{4x}}{x}}}{{\frac{{\sqrt {{x^2} + 2x + 4} }}{x} + \frac{{\sqrt {{x^2} - 2x + 4} }}{x}}}\\ = \mathop {\lim }\limits_{x \to  - \infty } \frac{4}{{ - \sqrt {1 + \frac{2}{x} + \frac{4}{{{x^2}}}}  - \sqrt {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} }} = \frac{4}{{ - 1 - 1}} =  - 2\end{array}\)

 

\( \Rightarrow \mathop {\lim }\limits_{x \to  + \infty } f(x) \ne \mathop {\lim }\limits_{x \to  - \infty } f(x)\)

Vậy không tồn tại \(\mathop {\lim }\limits_{x \to \infty } f(x)\).

Chọn: D.