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Câu hỏi: Cho $\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( x \right)-10}{x-1}=5$. Tính $\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( x \right)-10}{\left( \sqrt{x}-1 \right)\left( \sqrt{4f\left( x \right)+9}+3 \right)}$ ?
A. $1$.
B. $2$.
C. $10$.
D. $\dfrac{5}{3}$.
Khi đó $\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( x \right)-10}{\left( \sqrt{x}-1 \right)\left( \sqrt{4f\left( x \right)+9}+3 \right)}=\underset{x\to 1}{\mathop{\lim }} \left[ \dfrac{f\left( x \right)-10}{x-1}.\dfrac{\sqrt{x}+1}{\sqrt{4f\left( x \right)+9}+3} \right]=5.\dfrac{1+1}{\sqrt{4.10+9}+3}=1$.
A. $1$.
B. $2$.
C. $10$.
D. $\dfrac{5}{3}$.
Ta có $\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( x \right)-10}{x-1}=\underset{x\to {{x}_{0}}}{\mathop{\lim }} \dfrac{f(x)-f\left( {{x}_{0}} \right)}{x-{{x}_{0}}}={f}'\left( {{x}_{0}} \right)$ suy ra $f\left( 1 \right)=10$ và ${f}'\left( 1 \right)=5$.Khi đó $\underset{x\to 1}{\mathop{\lim }} \dfrac{f\left( x \right)-10}{\left( \sqrt{x}-1 \right)\left( \sqrt{4f\left( x \right)+9}+3 \right)}=\underset{x\to 1}{\mathop{\lim }} \left[ \dfrac{f\left( x \right)-10}{x-1}.\dfrac{\sqrt{x}+1}{\sqrt{4f\left( x \right)+9}+3} \right]=5.\dfrac{1+1}{\sqrt{4.10+9}+3}=1$.
Đáp án A.