Bài 2:
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{2x^2+x-3}{x-1}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(2x+3\right)\left(x-1\right)}{x-1}=\lim\limits_{x\rightarrow1^+}2x+3=2\cdot1+3=5\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}m-2=m-2\)
\(f\left(1\right)=m-2\)
Để hàm số f(x) liên tục tại x=1 thì \(\lim\limits_{x\rightarrow1^-}=\lim\limits_{x\rightarrow1^+}=f\left(1\right)\)
=>m-2=5
=>m=7
Bài 1:
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x}-1}{x^3-1}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\left(\sqrt{x}+1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\left(1+1\right)\left(1+1+1\right)}=\dfrac{1}{6}\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}m^2x^2+mx\)
\(=m^2\cdot\left(1\right)^2+m\cdot\left(1\right)=m^2+m\)
\(f\left(1\right)=m^2\cdot1^2+m\cdot1=m^2+m\)
Để hàm số có \(\lim\limits_{x\rightarrow1}f\left(x\right)\) thì \(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^+}f\left(x\right)\)
=>\(m^2+m=\dfrac{1}{6}\)
=>\(m^2+m-\dfrac{1}{6}=0\)
=>\(m=\dfrac{-3\pm\sqrt{15}}{6}\)
