Đáp án:
$\begin{array}{l}
y = \sqrt {{x^2} + x + 1} - \sqrt {{x^2} - x + 1} \\
= \dfrac{{{x^2} + x + 1 - {x^2} + x - 1}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x + 1} }}\\
= \dfrac{{2x}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x + 1} }}\\
Do:\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x + 1} > 0
\end{array}$
=> Hàm số ko có TCĐ
$\begin{array}{l}
y = \dfrac{{2x}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x + 1} }}\\
+ \mathop {\lim }\limits_{x \to + \infty } y = \mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x}}{{\sqrt {{x^2} + x + 1} + \sqrt {{x^2} - x + 1} }}\\
= \mathop {\lim }\limits_{x \to + \infty } \dfrac{2}{{\sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} + \sqrt {1 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} }}\\
= \dfrac{2}{{1 + 1}} = 1\\
+ )\mathop {\lim }\limits_{x \to - \infty } y\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{2}{{ - \sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} - \sqrt {1 - \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} }}\\
= \dfrac{2}{{ - 1 - 1}} = - 1
\end{array}$
=> Hàm số có 2 TCN là y=1 và y=-1
