Giải thích các bước giải:

\(\begin{array}{l}
e,\\
\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 3x + 2}}{{{{\left( {x - 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to 2} \frac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{{{\left( {x - 2} \right)}^2}}} = \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{{x - 2}}\\
\mathop {\lim }\limits_{x \to 2} \left( {x - 1} \right) = 2 - 1 = 1\\
\mathop {\lim }\limits_{x \to 2} \left( {x - 2} \right) = 2 - 2 = 0\\
 \Rightarrow \mathop {\lim }\limits_{x \to 2} \frac{{x - 1}}{{x - 2}} = \infty \\
h,\\
\mathop {\lim }\limits_{x \to 1} \frac{{4{x^6} - 5{x^5} + x}}{{{{\left( {1 - x} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x\left( {4{x^5} - 5{x^4} + 1} \right)}}{{{{\left( {1 - x} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x.\left[ {\left( {4{x^5} - 4{x^4}} \right) - \left( {{x^4} - {x^3}} \right) - \left( {{x^3} - {x^2}} \right) - \left( {{x^2} - x} \right) - \left( {x - 1} \right)} \right]}}{{{{\left( {1 - x} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x.\left( {x - 1} \right)\left( {4{x^4} - {x^3} - {x^2} - x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x\left( {x - 1} \right)\left[ {\left( {4{x^4} - 4{x^3}} \right) + \left( {3{x^3} - 3{x^2}} \right) + \left( {2{x^2} - 2x} \right) + \left( {x - 1} \right)} \right]}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{x\left( {x - 1} \right)\left( {x - 1} \right)\left( {4{x^3} + 3{x^2} + 2x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to 1} \left[ {x.\left( {4{x^3} + 3{x^2} + 2x + 1} \right)} \right]\\
 = 1.\left[ {{{4.1}^3} + {{3.1}^2} + 2.1 + 1} \right]\\
 = 10\\
i,\\
\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{8x + 11}} - \sqrt {x + 7} }}{{{x^2} - 3x + 2}}\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{\sqrt[3]{{8x + 11}} - 3}}{{{x^2} - 3x + 2}} + \frac{{3 - \sqrt {x + 7} }}{{{x^2} - 3x + 2}}} \right]\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{\left( {8x + 11} \right) - {3^3}}}{{\left( {{x^2} - 3x + 2} \right)\left( {{{\sqrt[3]{{8x + 11}}}^2} + 3.\sqrt[3]{{8x + 11}} + {3^2}} \right)}} + \frac{{{3^2} - \left( {x + 7} \right)}}{{\left( {{x^2} - 3x + 2} \right).\left( {3 + \sqrt {x + 7} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{{8\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {{{\sqrt[3]{{8x + 11}}}^2} + 3.\sqrt[3]{{8x + 11}} + {3^2}} \right)}} - \frac{{x - 2}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {3 + \sqrt {x + 7} } \right)}}} \right]\\
 = \mathop {\lim }\limits_{x \to 2} \left[ {\frac{8}{{\left( {x - 1} \right).\left( {{{\sqrt[3]{{8x + 11}}}^2} + 3.\sqrt[3]{{8x + 11}} + {3^2}} \right)}} - \frac{1}{{\left( {x - 1} \right)\left( {3 + \sqrt {x + 7} } \right)}}} \right]\\
 = \frac{8}{{\left( {2 - 1} \right).\left( {{{\sqrt[3]{{8.2 + 11}}}^2} + 3.\sqrt[3]{{8.2 + 11}} + {3^2}} \right)}} - \frac{1}{{\left( {2 - 1} \right).\left( {3 + \sqrt {2 + 7} } \right)}}\\
 = \frac{8}{{27}} - \frac{1}{6}\\
 = \frac{7}{{54}}
\end{array}\)