\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{2x+7}-3}{x-1}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{2x+7-9}{\sqrt{2x+7}+3}\cdot\dfrac{1}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{2}{\sqrt{2x+7}+3}\)
\(=\dfrac{2}{\sqrt{2+7}+3}=\dfrac{2}{6}=\dfrac{1}{3}\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}mx+5=m+5\)
f(1)=m*1+5=m+5
Để hàm số liên tục tại x=1 thì m+5=1/3
=>m=-14/3
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