Đáp án:

$\begin{array}{l}
1)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 3{x^2} + 2}}{{{x^3} + 2x - 3}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {{x^2} - 1} \right)\left( {{x^2} - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 3} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 1} \right)\left( {{x^2} - 2} \right)}}{{{x^2} + x + 3}}\\
 = \dfrac{{\left( {1 + 1} \right).\left( {{1^2} - 2} \right)}}{{{1^2} + 1 + 3}}\\
 = \dfrac{{ - 2}}{5}\\
2)\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {{x^2} - 3x + 4}  - 2}}{x}\\
 = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^2} - 3x + 4 - 4}}{{\left( {\sqrt {{x^2} - 3x + 4}  + 2} \right).x}}\\
 = \mathop {\lim }\limits_{x \to 0} \dfrac{{x - 3}}{{\sqrt {{x^2} - 3x + 4}  + 2}}\\
 = \dfrac{{0 - 3}}{{\sqrt {{0^2} - 3.0 + 4}  + 2}}\\
 = \dfrac{{ - 3}}{4}\\
3)\mathop {\lim }\limits_{x \to 2} \dfrac{{{x^2} - 5x + 6}}{{\sqrt {4x + 1}  - 3}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{{4x + 1 - 9}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{\left( {x - 3} \right)\left( {\sqrt {4x + 1}  + 3} \right)}}{4}\\
 = \dfrac{{\left( {2 - 3} \right).\left( {\sqrt {4.2 + 1}  + 3} \right)}}{4}\\
 = \dfrac{{ - 3}}{2}
\end{array}$