Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
\mathop {\lim }\limits_{x \to + \infty } = \frac{{x + \sqrt {x + \sqrt x } - x}}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} = \frac{{\sqrt {x + \sqrt x } }}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }}\\
\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } = \frac{{\sqrt {\frac{{x + \sqrt x }}{x}} }}{{\sqrt {\frac{{x + \sqrt {x + \sqrt x } }}{x}} + 1}} = \frac{{\sqrt {1 + \frac{1}{{\sqrt x }}} }}{{\sqrt {1 + \sqrt {\frac{{x + \sqrt x }}{{{x^2}}}} } + 1}}\\
\Leftrightarrow \mathop {\lim }\limits_{x \to + \infty } = \frac{{\sqrt {1 + \frac{1}{{\sqrt x }}} }}{{\sqrt {1 + \sqrt {\frac{1}{x} + \frac{1}{{\sqrt {{x^3}} }}} } + 1}} = \frac{{\sqrt {1 + \frac{1}{{ + \infty }}} }}{{\sqrt {1 + \sqrt {\frac{1}{{ + \infty }} + \frac{1}{{ + \infty }}} } + 1}}\\
= \frac{{\sqrt {1 + 0} }}{{\sqrt {1 + \sqrt {0 + 0} } + 1}} = \frac{1}{2}
\end{array}$
