Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + 2x + 5} + x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {\sqrt {{x^2} + 2x + 5} + x} \right)\left( {\sqrt {{x^2} + 2x + 5} - x} \right)}}{{\sqrt {{x^2} + 2x + 5} - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {{x^2} + 2x + 5} \right) - {x^2}}}{{\sqrt {{x^2}\left( {1 + \dfrac{2}{x} + \dfrac{5}{{{x^2}}}} \right)} - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x + 5}}{{\left| x \right|.\sqrt {1 + \dfrac{2}{x} + \dfrac{5}{{{x^2}}}} - x}}\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x + 5}}{{ - x.\sqrt {1 + \dfrac{2}{x} + \dfrac{5}{{{x^2}}}} - x}}\,\,\,\,\,\,\,\,\,\left( {x \to - \infty \Rightarrow x < 0 \Rightarrow \left| x \right| = - x} \right)\\
= \mathop {\lim }\limits_{x \to - \infty } \dfrac{{2 + \dfrac{5}{x}}}{{ - \sqrt {1 + \dfrac{2}{x} + \dfrac{5}{{{x^2}}}} - 1}}\\
= \dfrac{{2 + 0}}{{ - \sqrt {1 + 0 + 0} - 1}}\\
= - 1\\
2,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} + 2x + 5} + x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2}\left( {1 + \dfrac{2}{x} + \dfrac{5}{{{x^2}}}} \right)} + x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left( {x\sqrt {1 + \dfrac{2}{x} + \dfrac{5}{{{x^2}}}} + x} \right)\,\,\,\,\,\,\,\,\,\,\,\left( {x \to + \infty \Rightarrow x > 0} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \left[ {x.\left( {\sqrt {1 + \dfrac{2}{x} + \dfrac{5}{{{x^2}}}} + 1} \right)} \right]\\
= + \infty \\
\left( \begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } x = + \infty \\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {1 + \dfrac{2}{x} + \dfrac{5}{{{x^2}}}} + 1} \right) = 2
\end{array} \right)
\end{array}\)
