Cho f (x) là đa thức thỏa mãn \(\lim \limits_{x \rightarrow 3} \frac{f(x)-15}{x-3}=12 . \text { Tính } T=\lim\limits _{x \rightarrow 3} \frac{\sqrt[3]{5 f(x)-11}-4}{x^{2}-x-6}\)
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Lời giải:
Báo saiTa có
\(\begin{array}{l} \mathop {\lim }\limits_{x \to 3} \frac{{f(x) - 15}}{{x - 3}} = 12\\ \text{Đặt }f\left( x \right) = ax + b\\ \mathop {\lim }\limits_{x \to 3} \frac{{f(x) - 15}}{{x - 3}} = 12 \Leftrightarrow \frac{{f(x) - 15}}{{x - 3}} = 12 \Leftrightarrow \frac{{ax + b - 15}}{{x - 3}} = 12\\ \Leftrightarrow ax + b - 15 = 12\left( {x - 3} \right) \Leftrightarrow a\left( {x + \frac{{b - 15}}{a}} \right) = 12\left( {x - 3} \right)\\ \Leftrightarrow \left\{ \begin{array}{l} a = 12\\ \frac{{b - 15}}{a} = - 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 12\\ b = - 21 \end{array} \right.\\ \Rightarrow f\left( x \right) = 12x - 21 \end{array}\)
Khi đó ta có
\(\begin{array}{l} T = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt[3]{{5 \cdot f(x) - 11}} - 4}}{{{x^2} - x - 6}}\\ = \mathop {\lim }\limits_{x \to 3} \frac{{5f(x) - 75}}{{(x + 2)(x - 3) \cdot \left[ {{{\sqrt[3]{{5 \cdot f(x) - 11}}}^2} + 4 \cdot \sqrt[3]{{5.f(x) - 11}} + 16} \right]}}\\ = \mathop {\lim }\limits_{x \to 3} \frac{{5 \cdot (f(x) - 15)}}{{(x - 3)(x + 2)\left[ {{{\sqrt[3]{{5 \cdot f(x) - 11}}}^2} + 4 \cdot \sqrt[3]{{5.f(x) - 11}} + 16} \right]}}\\ = \mathop {\lim }\limits_{x \to 3} \frac{{f(x) - 15}}{{x - 3}} \cdot \mathop {\lim }\limits_{x \to 3} \frac{5}{{(x + 2)\left[ {{{\sqrt[3]{{5 \cdot f(x) - 11}}}^2} + 4 \cdot \sqrt[3]{{5.f(x) - 11}} + 16} \right]}}\\ = 12.\mathop {\lim }\limits_{x \to 3} \frac{5}{{(x + 2)\left[ {{{\sqrt[3]{{5 \cdot \left( {12x - 21} \right) - 11}}}^2} + 4 \cdot \sqrt[3]{{5.\left( {12x - 21} \right) - 11}} + 16} \right]}}\\ = 12.\frac{1}{{48}} = \frac{1}{4} \end{array}\)
