Giải thích các bước giải:
a)
$\begin{array}{l}
{I_5} = \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + x + 1} + x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2}\left( {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right)} + x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {x\sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} + x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {1 + \dfrac{1}{x} + \dfrac{1}{{{x^2}}}} + 1} \right)\\
= + \infty
\end{array}$
b)
$\begin{array}{l}
{I_6} = \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + x + 2} + x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2}\left( {1 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}}} \right)} + x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \left( { - x\sqrt {1 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}}} + x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } x\left( {1 - \sqrt {1 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}}} } \right)\\
= \mathop {\lim }\limits_{x \to - \infty } x.\dfrac{{1 - \left( {1 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}}} \right)}}{{1 + \sqrt {1 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}}} }}\\
= \mathop {\lim }\limits_{x \to - \infty } x.\dfrac{{\dfrac{{ - 1}}{x} - \dfrac{1}{{{x^2}}}}}{{1 + \sqrt {1 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}}} }}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{ - 1 - \dfrac{1}{x}}}{{1 + \sqrt {1 + \dfrac{1}{x} + \dfrac{2}{{{x^2}}}} }}\\
= \dfrac{{ - 1 - 0}}{{1 + \sqrt {1 + 0 + 2.0} }}\\
= \dfrac{{ - 1}}{2}
\end{array}$
