Giải thích các bước giải:
\(\begin{array}{l}
27,\\
\lim \frac{{2 - {5^{n - 2}}}}{{{3^n} + {{2.5}^n}}}\\
= \lim \frac{{2.{{\left( {\frac{1}{5}} \right)}^n} - \frac{{{5^{n - 2}}}}{{{5^n}}}}}{{{{\left( {\frac{3}{5}} \right)}^n} + 2}}\\
= \lim \frac{{2.{{\left( {\frac{1}{5}} \right)}^n} - \frac{1}{{25}}}}{{{{\left( {\frac{3}{5}} \right)}^n} + 2}}\\
= \frac{{2.0 - \frac{1}{{25}}}}{{0 + 2}}\\
= - \frac{1}{{50}}\\
28,\\
\lim \frac{{{3^n} - {{4.2}^{n - 1}} - 3}}{{{{3.2}^n} + {4^n}}}\\
= \lim \frac{{{3^n} - {{2.2}^n} - 3}}{{{{3.2}^n} + {4^n}}}\\
= \lim \frac{{{{\left( {\frac{3}{4}} \right)}^n} - 2.{{\left( {\frac{1}{2}} \right)}^n} - 3.{{\left( {\frac{1}{4}} \right)}^n}}}{{3.{{\left( {\frac{1}{2}} \right)}^n} + 1}}\\
= \frac{{0 - 2.0 - 3.0}}{{3.0 + 1}} = 0\\
29,\\
\mathop {\lim }\limits_{x \to + \infty } \left( {\sqrt {{x^2} - x + 1} - x} \right)\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} - x + 1 - {x^2}}}{{\sqrt {{x^2} - x + 1} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - x + 1}}{{\sqrt {{x^2} - x + 1} + x}}\\
= \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 1 + \frac{1}{x}}}{{\sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}} + 1}}\\
= \frac{{ - 1}}{{\sqrt 1 + 1}}\\
= - \frac{1}{2}
\end{array}\)
