Đáp án:

$\begin{array}{l}
1)\mathop {\lim }\limits_{x \to {{\left( { - \dfrac{1}{4}} \right)}^ + }} \dfrac{{2x - 5}}{{4x + 1}} = \dfrac{{2.\dfrac{{ - 1}}{4} - 5}}{{4.\dfrac{{ - 1}}{4} + 1}} =  - \infty \\
2)\mathop {\lim }\limits_{x \to 2} \dfrac{{\sqrt {2{x^2} + 1}  - 2x + 1}}{{8 - 4x}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{2{x^2} + 1 - 4{x^2} - 4x - 1}}{{ - 4\left( {x - 2} \right)\left( {\sqrt {2{x^2} + 1}  + 2x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{{ - 2\left( {x - 2} \right)}}{{ - 4\left( {x - 2} \right)\left( {\sqrt {2{x^2} + 1}  + 2x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 2} \dfrac{1}{{2\left( {\sqrt {2{x^2} + 1}  + 2x - 1} \right)}}\\
 = \dfrac{1}{{2\left( {\sqrt {{{2.2}^2} + 1}  + 2.2 - 1} \right)}}\\
 = \dfrac{1}{{12}}\\
3)\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {2{x^2} + 1}  - 2x + 1}}{{8 - 4x}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{\sqrt {2 + \dfrac{1}{{{x^2}}}}  - 2 + \dfrac{1}{x}}}{{\dfrac{8}{x} - 4}}\\
 = \dfrac{{\sqrt 2  - 2}}{{ - 4}} = \dfrac{{2 - \sqrt 2 }}{4}\\
4)\mathop {\lim }\limits_{x \to 1} \dfrac{{3{x^4} + {x^3} - 5x + 1}}{{1 - x}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {3{x^3} + 4{x^2} + 4x - 1} \right)}}{{1 - x}}\\
 = \mathop {\lim }\limits_{x \to 1} \left( {1 - 4x - 4{x^2} - 3{x^3}} \right)\\
 = 1 - 4.1 - {4.1^2} - {3.1^3}\\
 =  - 10\\
5)\mathop {\lim }\limits_{x \to  + \infty } \dfrac{{3{x^4} + {x^3} - 5x + 1}}{{1 - x}}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \left( {1 - 4x - 4{x^2} - 3{x^3}} \right)\\
 =  - \infty \\
6)\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{3{x^4} + {x^3} - 5x + 1}}{{1 - x}}\\
 = \mathop {\lim }\limits_{x \to {1^ - }} \left( {1 - 4x - 4{x^2} - 3{x^3}} \right)\\
 =  - 10\\
7)\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x - cos4x}}{{4co{s^2}x}}\\
 = \dfrac{{\sin 0 - c{\rm{os0}}}}{{4,co{s^2}0}} = \dfrac{{ - 1}}{4}\\
8)\mathop {\lim }\limits_{x \to 1} \dfrac{{4x - 5}}{{{{\left( {x - 1} \right)}^2}}} =  - \infty \\
9)\mathop {\lim }\limits_{x \to {2^ - }} \dfrac{{4{x^3} + {x^2} - 5}}{{2x - 4}}\\
 = \dfrac{{{{4.2}^3} + {2^2} - 5}}{{2.2 - 4}} =  - \infty \\
10)\mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {2x + 7}  - 3x}}{{2x - \sqrt {3{x^2} + 1} }}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\sqrt {2x + 7}  - 3 + 3 - 3x}}{{2x - 2 + 2 - \sqrt {3{x^2} + 1} }}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{2x + 7 - 9}}{{\sqrt {2x + 7}  + 3}} - 3\left( {x - 1} \right)}}{{2\left( {x - 1} \right) + \dfrac{{4 - 3{x^2} - 1}}{{2 + \sqrt {3{x^2} + 1} }}}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{2}{{\sqrt {2x + 7}  + 3}} - 3}}{{2 + \dfrac{{ - 3\left( {x + 1} \right)}}{{2 + \sqrt {3{x^2} + 1} }}}}\\
 = \dfrac{{\dfrac{2}{{\sqrt {2.1 + 7}  + 3}} - 3}}{{2 + \dfrac{{ - 3.\left( {1 + 1} \right)}}{{2 + \sqrt {3.1 + 1} }}}} = \dfrac{{\dfrac{1}{3} - 3}}{{2 - \dfrac{3}{2}}} =  - \dfrac{{16}}{3}
\end{array}$