Cho $f(x)$ là đa thức thỏa mãn $\lim_{x \to 1} \frac{f(x) - 5}{x - 1} = 10$. Gọi $T = \lim_{x \to 1} \frac{\sqrt[3]{4f(x) + 7} - \sqrt{f(x) + 4}}{x^{2} + x - 2}$. Tính $81T$.

Lời giải Đặt $g(x) = \frac{f(x) - 5}{x - 1} \Rightarrow f(x) = (x - 1)g(x) + 5 \Rightarrow \lim_{x \to 1} f(x) = 5$. Ta có: $$ T = \lim_{x \to 1} \frac{\sqrt[3]{4f(x) + 7} - \sqrt{f(x) + 4}}{x^{2} + x - 2} = \lim_{x \to 1} \frac{\Big(\sqrt[3]{4f(x) + 7} - 3\Big) + \Big(3 - \sqrt{f(x) + 4}\Big)}{x^{2} + x - 2} $$ $$ = \lim_{x \to 1} \Big( \frac{\sqrt[3]{4f(x) + 7} - 3}{x^{2} + x - 2} - \frac{\sqrt{f(x) + 4} - 3}{x^{2} + x - 2} \Big) $$ $$ = \lim_{x \to 1} \Big( \frac{4[f(x) - 5]}{(x - 1)(x + 2)\Big[ \sqrt[3]{(4f(x) + 7)^{2}} + 3\sqrt[3]{4f(x) + 7} + 9 \Big]} - \frac{f(x) - 5}{(x - 1)(x + 2)\Big[ \sqrt{f(x) + 4} + 3 \Big]} \Big) $$ $$ = \lim_{x \to 1} \left[ \frac{f(x) - 5}{x - 1} \cdot \Big( \frac{4}{(x + 2)\Big( \sqrt[3]{(4f(x) + 7)^{2}} + 3\sqrt[3]{4f(x) + 7} + 9 \Big)} - \frac{1}{(x + 2)\Big( \sqrt{f(x) + 4} + 3 \Big)} \Big) \right] $$ $$ = 10 \cdot \Big( \frac{4}{81} - \frac{1}{18} \Big) = -\frac{5}{81} \Rightarrow 81T = -5 $$