Đáp án:
$\begin{array}{l}
14)\mathop {\lim }\limits_{x \to {3^ + }} \frac{{x - 3}}{{\sqrt {{x^2} - 9} }} = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{{{\left( {\sqrt {x - 3} } \right)}^2}}}{{\sqrt {x - 3} .\sqrt {x + 3} }}\\
= \mathop {\lim }\limits_{x \to {3^ + }} \frac{{\sqrt {x - 3} }}{{\sqrt {x + 3} }} = 0\\
16)\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 3x + 2}}{{{x^3} - {x^2} + x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + 1} \right)}}\\
= \mathop {\lim }\limits_{x \to 1} \frac{{x - 2}}{{{x^2} + 1}}\\
= \frac{{1 - 2}}{{1 + 1}} = \frac{{ - 1}}{2}
\end{array}$
