Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left| {3 - x} \right|}}{{3 - x}}\\
x \to {3^ + } \Rightarrow x > 3 \Rightarrow \left| {3 - x} \right| = x - 3\\
\Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\left| {3 - x} \right|}}{{3 - x}} = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 3}}{{3 - x}} = - 1\\
2,\\
\mathop {\lim }\limits_{x \to {0^ + }} \frac{{x + 2\sqrt x }}{{x - \sqrt x }}\\
= \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt x + 2}}{{\sqrt x - 1}}\\
= \frac{{\sqrt 0 + 2}}{{\sqrt 0 - 1}}\\
= - 2
\end{array}\)
