Cho \(f\left( x \right)\) là đa thức thỏa mãn \(\mathop {\lim }\limits_{x \to 3} \dfrac{{f\left( x \right) - 8}}{{x - 3}} = 6\). Tính \(L = \mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt[3]{{f\left( x \right) - 7}} - 1}}{{{x^2} - 2x - 3}}\).

Đáp án: 

Bước 1: Tìm \(\mathop {\lim }\limits_{x \to 3} f\left( x \right)\).

Ta thấy: \(\mathop {\lim }\limits_{x \to 3} \dfrac{{f\left( x \right) - 8}}{{x - 3}} = 6\) 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 \(\mathop {\lim }\limits_{x \to 3} \left[ {f\left( x \right) - 8} \right] \ne 0,\,\,\mathop {\lim }\limits_{x \to 3} \left( {x - 3} \right) = 0 \Rightarrow \mathop {\lim }\limits_{x \to 3} \dfrac{{f\left( x \right) - 8}}{{x - 3}} = \infty \)).

Bước 2: Biến đổi, làm mất dạng vô định để tìm giới hạn của hàm số.

Ta có:

\(\begin{array}{l}L = \mathop {\lim }\limits_{x \to 3} \dfrac{{\sqrt[3]{{f\left( x \right) - 7}} - 1}}{{{x^2} - 2x - 3}}\\ = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{{\left( {\sqrt[3]{{f\left( x \right) - 7}} - 1} \right)\left( {{{\sqrt[3]{{f\left( x \right) - 7}}}^2} + \sqrt[3]{{f\left( x \right) - 7}} + 1} \right)}}{{\left( {{{\sqrt[3]{{f\left( x \right) - 7}}}^2} + \sqrt[3]{{f\left( x \right) - 7}} + 1} \right)}}}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\end{array}\)

\(\begin{array}{l} = \mathop {\lim }\limits_{x \to 3} \dfrac{{\dfrac{{f\left( x \right) - 8}}{{\left( {{{\sqrt[3]{{f\left( x \right) - 7}}}^2} + \sqrt[3]{{f\left( x \right) - 7}} + 1} \right)}}}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\ = \mathop {\lim }\limits_{x \to 3} \left[ {\dfrac{{f\left( x \right) - 8}}{{x - 3}}.\dfrac{1}{{\left( {{{\sqrt[3]{{f\left( x \right) - 7}}}^2} + \sqrt[3]{{f\left( x \right) - 7}} + 1} \right)\left( {x + 1} \right)}}} \right]\\ = 6.\dfrac{1}{{\left( {{{\sqrt[3]{{8 - 7}}}^2} + \sqrt[3]{{8 - 7}} + 1} \right)\left( {3 + 1} \right)}} = \dfrac{6}{{3.4}} = \dfrac{1}{2}\end{array}\)