Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
223,\\
\lim \frac{{{n^3} + {n^2} - n}}{{1 - 2{n^2} - 4{n^3}}} = \lim \frac{{1 + \frac{1}{n} - \frac{1}{{{n^2}}}}}{{\frac{1}{{{n^3}}} - \frac{2}{n} - 4}} = \frac{{1 + 0 - 0}}{{0 - 0 - 4}} =  - \frac{1}{4}\\
224,\\
\lim \frac{{{{3.2}^{2n + 1}} + 3}}{{{{3.2}^n} + {{7.3}^n}}} = \lim \frac{{{{3.2.2}^{2n}} + 3}}{{{{3.2}^n} + {{7.3}^n}}} = \lim \frac{{{{6.4}^n} + 3}}{{{{3.2}^n} + {{7.3}^n}}} = \lim \frac{{6.{{\left( {\frac{4}{3}} \right)}^n} + {{\left( {\frac{1}{3}} \right)}^{n - 1}}}}{{3.{{\left( {\frac{2}{3}} \right)}^n} + 7}} = \frac{{ + \infty }}{7} =  + \infty \\
225,\\
\mathop {\lim }\limits_{x \to  + \infty } \frac{{a\,x + \sqrt {{x^2} - 3x + 5} }}{{2x - 7}} = 2\\
 \Leftrightarrow \mathop {\lim }\limits_{x \to  + \infty } \frac{{a + \sqrt {1 - \frac{3}{x} + \frac{5}{{{x^2}}}} }}{{2 - \frac{7}{x}}} = 2\\
 \Leftrightarrow \frac{{a + \sqrt 1 }}{2} = 2\\
 \Leftrightarrow a = 3\\
226,\\
\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\\
\mathop {\lim }\limits_{x \to 0} \frac{{\cos 3x - \cos 7x}}{{{x^2}}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2.\sin \frac{{3x + 7x}}{2}.\sin \frac{{3x - 7x}}{2}}}{{{x^2}}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2.\sin 5x.\sin \left( { - 2x} \right)}}{{{x^2}}}\\
 = \mathop {\lim }\limits_{x \to 0} \frac{{2\sin 5x.\sin 2x}}{{{x^2}}}\\
 = \mathop {\lim }\limits_{x \to 0} \left[ {20.\frac{{\sin 5x}}{{5x}}.\frac{{\sin 2x}}{{2x}}} \right] = 20.1.1 = 20
\end{array}\)