Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\frac{{9 - {x^2}}}{{x - 3}}\;\;{\rm{khi}}\;x < 3\\1 - x\;\;\;\;\;{\rm{khi}}\;x \ge 3\end{array} \right.\). Biết \(\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = a,\mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right) = b\). Tính \({a^2} + {b^2}\).

Từ giả thiết ta có \(\mathop {\lim }\limits_{x \to 2} \left( {f\left( x \right) - 20} \right) = 0 \Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = 20\).

\(T = \mathop {\lim }\limits_{x \to 2} \frac{{\sqrt[3]{{6f\left( x \right) + 5}} - 5}}{{{x^2} + x - 6}}\)\( = \mathop {\lim }\limits_{x \to 2} \frac{{6f\left( x \right) + 5 - 125}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\)

\[ = \mathop {\lim }\limits_{x \to 2} \frac{{6\left[ {f\left( x \right) - 20} \right]}}{{\left( {x - 2} \right)\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\]\[ = \mathop {\lim }\limits_{x \to 2} \frac{{6\left[ {f\left( x \right) - 20} \right]}}{{\left( {x - 2} \right)}} \cdot \mathop {\lim }\limits_{x \to 2} \frac{1}{{\left( {x + 3} \right)\left[ {{{\left( {\sqrt[3]{{6f\left( x \right) + 5}}} \right)}^2} + 5\sqrt[3]{{6f\left( x \right) + 5}} + 25} \right]}}\]

\[ = 6 \cdot 10 \cdot \frac{1}{{5 \cdot \left( {25 + 25 + 25} \right)}} \approx 0,2\].

Trả lời: 0,2.

a) \(\lim \frac{{{n^2} + 5n}}{{3{n^2} - 2n + 1}}\)\( = \lim \frac{{1 + \frac{5}{n}}}{{3 - \frac{2}{n} + \frac{1}{{{n^2}}}}} = \frac{1}{3}\).

b) \(\lim \frac{{{3^n} - 2 \cdot {4^n}}}{{5 \cdot {4^n} + {3^n}}}\)\( = \lim \frac{{{{\left( {\frac{3}{4}} \right)}^n} - 2}}{{5 + {{\left( {\frac{3}{4}} \right)}^n}}} = - \frac{2}{5}\).