Cho hàm số f ( x ) = căn bậc hai của x^2 + 2 x + 4 − căn bậc hai của x^2 − 2 x + 4 . Khẳng định nào sau đây là đúng?

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

Ta có:

\[\mathop {lim}\limits_{x \to + \infty } f(x) = \mathop {lim}\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} } \right)\left( {\sqrt {{x^2} + 2x + 4} - \sqrt {{x^2} - 2x + 4} } \right)\]

\[ = \mathop {lim}\limits_{x \to + \infty } \frac{{\left( {\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} } \right)}}{{\left( {\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} } \right)}}\]

\[ = \mathop {lim}\limits_{x \to + \infty } \frac{{({x^2} + 2x + 4) - ({x^2} - 2x + 4)}}{{\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} }}\]

\[\begin{array}{l} = \mathop {lim}\limits_{x \to + \infty } \frac{{4x}}{{\sqrt {{x^2} + 2x + 4} + \sqrt {{x^2} - 2x + 4} }}\\ = \mathop {lim}\limits_{x \to + \infty } \frac{4}{{\sqrt {1 + \frac{2}{x} + \frac{4}{{{x^2}}}} + \sqrt {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} }} = 2\end{array}\]

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

\[ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } f(x) = - \mathop {\lim }\limits_{x \to - \infty } f(x)\]