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Câu hỏi: Cho f( x) là đa thức thỏa mãn $\underset{x\to 3}{\mathop{\lim }} \dfrac{f\left( x \right)-8}{x-3}=6$. Tính $L=\underset{x\to 3}{\mathop{\lim }} \dfrac{\sqrt[3]{f\left( x \right)-7}-1}{{{x}^{2}}-2x-3}$
A. $L=\dfrac{3}{4}$
B. $L=\dfrac{3}{2}$
C. $L=\dfrac{1}{2}$
D. $L=\dfrac{1}{4}$
A. $L=\dfrac{3}{4}$
B. $L=\dfrac{3}{2}$
C. $L=\dfrac{1}{2}$
D. $L=\dfrac{1}{4}$
Phương pháp:
- Tìm $\underset{{{_{x}}_{\to 3}}}{\mathop{lim}} f\left( x \right).~$
- Biến đổi, làm mất dạng vô định để tìm giới hạn của hàm số.
Cách giải:
Ta thấy: $\underset{x~\to ~3}{\mathop{\lim }} \dfrac{f\left( x \right)-8}{x-3}=6$ nên $\underset{x\to 3}{\mathop{\lim }} \left[ \left( f\left( x \right) \right)-8 \right]=0\Leftrightarrow \underset{x\to 3}{\mathop{\lim }} f\left( x \right)=8\Rightarrow f\left( 3 \right)=8$
(Bởi vì nếu $\underset{x\to 3}{\mathop{\lim }} \left[ \left( f\left( x \right) \right)-8 \right]\ne 0,\underset{x\to 3}{\mathop{\lim }} \left( x-3 \right)=0\Rightarrow \underset{x\to 3}{\mathop{\lim }} \dfrac{f\left( x \right)-8}{x-3}=\infty $ )
Ta có:
$\begin{aligned}
& L=\underset{x\to 3}{\mathop{\lim }} \dfrac{\sqrt[3]{f\left( x \right)-7}-1}{{{x}^{2}}-2x-3} \\
& =\underset{x\to 3}{\mathop{\lim }} \dfrac{\dfrac{\left( \sqrt[3]{f\left( x \right)-7}-1 \right)\left( {{\sqrt[3]{f\left( x \right)-7}}^{2}} \right)+\sqrt[3]{f\left( x \right)-7}+1}{\left( {{\sqrt[3]{f\left( x \right)-7}}^{2}}+\sqrt[3]{f\left( x \right)-7}+1 \right)}}{\left( x+1 \right)\left( x-3 \right)} \\
& =\underset{x\to 3}{\mathop{\lim }} \dfrac{\dfrac{f\left( x \right)-8}{\left( {{\sqrt[3]{f\left( x \right)-7}}^{2}}+\sqrt[3]{f\left( x \right)-7}+1 \right)}}{\left( x+1 \right)\left( x-3 \right)} \\
& =\underset{x\to 3}{\mathop{\lim }} \left[ \dfrac{f\left( x \right)-8}{x-3}.\dfrac{1}{\left( {{\sqrt[3]{f\left( x \right)-7}}^{2}}+\sqrt[3]{f\left( x \right)-7}+1 \right)\left( x+1 \right)} \right] \\
\end{aligned}$
$=6.\dfrac{1}{\left( {{\sqrt[3]{8-7}}^{2}}+\sqrt[3]{8-7}+1 \right)\left( 3+1 \right)}=\dfrac{6}{3.4}=\dfrac{1}{2}$
- Tìm $\underset{{{_{x}}_{\to 3}}}{\mathop{lim}} f\left( x \right).~$
- Biến đổi, làm mất dạng vô định để tìm giới hạn của hàm số.
Cách giải:
Ta thấy: $\underset{x~\to ~3}{\mathop{\lim }} \dfrac{f\left( x \right)-8}{x-3}=6$ nên $\underset{x\to 3}{\mathop{\lim }} \left[ \left( f\left( x \right) \right)-8 \right]=0\Leftrightarrow \underset{x\to 3}{\mathop{\lim }} f\left( x \right)=8\Rightarrow f\left( 3 \right)=8$
(Bởi vì nếu $\underset{x\to 3}{\mathop{\lim }} \left[ \left( f\left( x \right) \right)-8 \right]\ne 0,\underset{x\to 3}{\mathop{\lim }} \left( x-3 \right)=0\Rightarrow \underset{x\to 3}{\mathop{\lim }} \dfrac{f\left( x \right)-8}{x-3}=\infty $ )
Ta có:
$\begin{aligned}
& L=\underset{x\to 3}{\mathop{\lim }} \dfrac{\sqrt[3]{f\left( x \right)-7}-1}{{{x}^{2}}-2x-3} \\
& =\underset{x\to 3}{\mathop{\lim }} \dfrac{\dfrac{\left( \sqrt[3]{f\left( x \right)-7}-1 \right)\left( {{\sqrt[3]{f\left( x \right)-7}}^{2}} \right)+\sqrt[3]{f\left( x \right)-7}+1}{\left( {{\sqrt[3]{f\left( x \right)-7}}^{2}}+\sqrt[3]{f\left( x \right)-7}+1 \right)}}{\left( x+1 \right)\left( x-3 \right)} \\
& =\underset{x\to 3}{\mathop{\lim }} \dfrac{\dfrac{f\left( x \right)-8}{\left( {{\sqrt[3]{f\left( x \right)-7}}^{2}}+\sqrt[3]{f\left( x \right)-7}+1 \right)}}{\left( x+1 \right)\left( x-3 \right)} \\
& =\underset{x\to 3}{\mathop{\lim }} \left[ \dfrac{f\left( x \right)-8}{x-3}.\dfrac{1}{\left( {{\sqrt[3]{f\left( x \right)-7}}^{2}}+\sqrt[3]{f\left( x \right)-7}+1 \right)\left( x+1 \right)} \right] \\
\end{aligned}$
$=6.\dfrac{1}{\left( {{\sqrt[3]{8-7}}^{2}}+\sqrt[3]{8-7}+1 \right)\left( 3+1 \right)}=\dfrac{6}{3.4}=\dfrac{1}{2}$
Đáp án C.