Google
Câu hỏi: Tìm $m$ để $\underset{x\to 1}{\mathop{\lim }} \dfrac{{{x}^{2}}+5x+m}{x-1}=7$
A. $4$.
B. $-6$.
C. $0$.
D. $2$.
A. $4$.
B. $-6$.
C. $0$.
D. $2$.
Ta có $\underset{x\to 1}{\mathop{\lim }} \dfrac{{{x}^{2}}+5x+m}{x-1}=\underset{x\to 1}{\mathop{\lim }} \left( x+\dfrac{6x+m}{x-1} \right)=1+\underset{x\to 1}{\mathop{\lim }} \dfrac{6x+m}{x-1}=1+6.\underset{x\to 1}{\mathop{\lim }} \dfrac{x+\dfrac{m}{6}}{x-1}$.
Khi đó $1+6.\underset{x\to 1}{\mathop{\lim }} \dfrac{x+\dfrac{m}{6}}{x-1}=7\Leftrightarrow \underset{x\to 1}{\mathop{\lim }} \dfrac{x+\dfrac{m}{6}}{x-1}=1\Rightarrow x+\dfrac{m}{6}=x-1\Leftrightarrow \dfrac{m}{6}=-1\Leftrightarrow m=-6$.
Khi đó $1+6.\underset{x\to 1}{\mathop{\lim }} \dfrac{x+\dfrac{m}{6}}{x-1}=7\Leftrightarrow \underset{x\to 1}{\mathop{\lim }} \dfrac{x+\dfrac{m}{6}}{x-1}=1\Rightarrow x+\dfrac{m}{6}=x-1\Leftrightarrow \dfrac{m}{6}=-1\Leftrightarrow m=-6$.
Đáp án B.