Cho \(f\left( x \right)\) là đa thức thỏa mãn \(\underset{x\to 3}{\mathop{\lim }}\,\frac{f\left( x \right)-8}{x-3}=6\).

* Hướng dẫn giải

Do \(\underset{x\to 3}{\mathop{\lim }}\,\frac{f\left( x \right)-8}{x-3}=6\) nên ta có \(f\left( 3 \right)=8\)

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

\(=\underset{x\to 3}{\mathop{\lim }}\,\left[ \frac{f\left( x \right)-8}{\left( x-3 \right)}.\frac{1}{\left( x+1 \right)\left( \sqrt[3]{{{\left( f\left( x \right)-7 \right)}^{2}}}+\sqrt[3]{f\left( x \right)-7}+1 \right)} \right]=6.\frac{1}{4.3}=\frac{1}{2}\).