Giải thích các bước giải:

\(\begin{array}{l}
o,\\
\mathop {\lim }\limits_{x \to 1} \frac{{x - 5{x^5} + 4{x^6}}}{{{{\left( {1 - x} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - {x^5}} \right) + 4\left( {{x^6} - {x^5}} \right)}}{{{{\left( {1 - x} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x\left( {1 - {x^4}} \right) + 4.{x^5}\left( {x - 1} \right)}}{{{{\left( {1 - x} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x.\left( {1 - x} \right).\left( {1 + x} \right).\left( {1 + {x^2}} \right) + 4{x^5}\left( {x - 1} \right)}}{{{{\left( {1 - x} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x.\left( {1 + x} \right).\left( {1 + {x^2}} \right) - 4{x^5}}}{{1 - x}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + {x^2}} \right)\left( {1 + {x^2}} \right) - 4{x^5}}}{{1 - x}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{4{x^5} - {x^4} - {x^3} - {x^2} - x}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {4{x^5} - 4{x^4}} \right) + \left( {3{x^4} - 3{x^3}} \right) + \left( {2{x^3} - 2{x^2}} \right) + \left( {{x^2} - x} \right)}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \left( {4{x^4} + 3{x^3} + 2{x^2} + x} \right)\\
 = 4 + 3 + 2 + 1 = 10\\
p,\\
\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^3} - {x^3}}}{h}\\
 = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h - x} \right).\left[ {{{\left( {x + h} \right)}^2} + \left( {x + h} \right).x + {x^2}} \right]}}{h}\\
 = \mathop {\lim }\limits_{h \to 0} \left[ {{{\left( {x + h} \right)}^2} + \left( {x + h} \right).x + {x^2}} \right]\\
 = 3{x^2}\\
q,\\
\mathop {\lim }\limits_{x \to a} \frac{{{x^2} - \left( {a + 1} \right)x + a}}{{{x^3} - {a^3}}}\\
 = \mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^2} - x} \right) - a\left( {x - 1} \right)}}{{\left( {x - a} \right)\left( {{x^2} + x.a + {a^2}} \right)}}\\
 = \mathop {\lim }\limits_{x \to a} \frac{{\left( {x - a} \right)\left( {x - 1} \right)}}{{\left( {x - a} \right)\left( {{x^2} + x.a + {a^2}} \right)}}\\
 = \mathop {\lim }\limits_{x \to a} \frac{{x - 1}}{{{x^2} + x.a + {a^2}}}\\
 = \frac{{a - 1}}{{3{a^2}}}\\
r,\\
\mathop {\lim }\limits_{x \to a} \frac{{{x^4} - {a^4}}}{{x - a}}\\
 = \mathop {\lim }\limits_{x \to a} \frac{{\left( {x - a} \right)\left( {x + a} \right)\left( {{x^2} + {a^2}} \right)}}{{\left( {x - a} \right)}}\\
 = \mathop {\lim }\limits_{x \to a} \left[ {\left( {x + a} \right).\left( {{x^2} + {a^2}} \right)} \right]\\
 = \left( {a + a} \right).\left( {{a^2} + {a^2}} \right)\\
 = 2a.2{a^2} = 4{a^3}
\end{array}\)