Cho f( x ) là đa thức thỏa mãn lim x đến 3 (f(x)-8)/(x-3)=6. Tính

Đáp án: \(L = \frac{1}{2}\)

Phương pháp giải:

- Tìm \(\mathop {\lim }\limits_{x \to 3} f\left( x \right)\).

- Biến đổi, làm mất dạng vô định để tìm giới hạn của hàm số.

Giải chi tiết:

Ta thấy: \(\mathop {\lim }\limits_{x \to 3} \frac{{f\left( x \right) - 8}}{{x - 3}} = 6\) nên \(\mathop {\lim }\limits_{x \to 3} \left[ {\left( {f\left( x \right)} \right) - 8} \right] = 0 \Leftrightarrow \mathop {\lim }\limits_{x \to 3} f\left( x \right) = 8\)\( \Rightarrow f\left( 3 \right) = 8.\)

(Bởi vì nếu \(\mathop {\lim }\limits_{x \to 3} \left[ {f\left( x \right) - 8} \right] \ne 0,{\mkern 1mu} {\mkern 1mu} \mathop {\lim }\limits_{x \to 3} \left( {x - 3} \right) = 0 \Rightarrow \mathop {\lim }\limits_{x \to 3} \frac{{f\left( x \right) - 8}}{{x - 3}} = \infty \)).

Ta có: \[L = \mathop {\lim }\limits_{x \to 3} \frac{{\sqrt[3]{{f\left( x \right) - 7}} - 1}}{{{x^2} - 2x - 3}}\]

\[ = \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{\left( {\sqrt[3]{{f\left( x \right) - 7}} - 1} \right)\left( {{{\sqrt[3]{{f\left( x \right) - 7}}}^2} + \sqrt[3]{{f\left( x \right) - 7}} + 1} \right)}}{{\left( {{{\sqrt[3]{{f\left( x \right) - 7}}}^2} + \sqrt[3]{{f\left( x \right) - 7}} + 1} \right)}}}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\]

\[ = \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{f\left( x \right) - 8}}{{\left( {{{\sqrt[3]{{f\left( x \right) - 7}}}^2} + \sqrt[3]{{f\left( x \right) - 7}} + 1} \right)}}}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\]

\[ = \mathop {\lim }\limits_{x \to 3} \left[ {\frac{{f\left( x \right) - 8}}{{x - 3}}.\frac{1}{{\left( {{{\sqrt[3]{{f\left( x \right) - 7}}}^2} + \sqrt[3]{{f\left( x \right) - 7}} + 1} \right)\left( {x + 1} \right)}}} \right]\]

\[ = 6.\frac{1}{{\left( {{{\sqrt[3]{{8 - 7}}}^2} + \sqrt[3]{{8 - 7}} + 1} \right)\left( {3 + 1} \right)}} = \frac{6}{{3.4}} = \frac{1}{2}\].