\(f\left(-1\right)=\lim\limits_{x\rightarrow-1^-}f\left(x\right)=\lim\limits_{x\rightarrow-1^-}\left(2-ax\right)=2+a\)

\(\lim\limits_{x\rightarrow-1^+}f\left(x\right)=\lim\limits_{x\rightarrow-1^+}\left(x^2-bx+2\right)=3+b\)

\(f\left(1\right)=\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\left(4x+a\right)=4+a\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(x^2-bx+2\right)=3-b\)

Hàm liên tục trên R khi và chỉ khi:

\(\left\{{}\begin{matrix}2+a=3+b\\4+a=3-b\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=-1\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)

\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(ax+2\right)=a+2\)

Hàm liên tục tại x=1 khi:

\(a+2=\dfrac{1}{4}\Rightarrow a=-\dfrac{7}{4}\)

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}x^2+3x+1=1+3\cdot1+1=5\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}2x+2=2\cdot1+2=4\)

f(1)=1+3+1=5

=>\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=f\left(1\right)\ne\lim\limits_{x\rightarrow1^-}f\left(x\right)\)

=>Hàm số bị gián đoạn tại x=1

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)

\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(mx\right)=m\)

Hàm liên tục tại x=1 khi: \(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\)

\(\Leftrightarrow m=\dfrac{1}{4}\)

\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{x^3-x^2+2x-2}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{x^2\left(x-1\right)+2\left(x-1\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2+2\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(x^2+2\right)=3\)

\(f\left(1\right)=3.1+m=m+3\)

Hàm số liên tục tại \(x_0=1\) khi và chỉ khi \(\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\)

\(\Rightarrow m+3=3\Rightarrow m=0\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{ax+1}-\sqrt[]{1-bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\dfrac{bx}{1+\sqrt[]{1-bx}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{a}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\dfrac{b}{1+\sqrt[]{1-bx}}\right)=\dfrac{a}{3}+\dfrac{b}{2}\)

Hàm liên tục tại \(x=0\) khi:

\(\dfrac{a}{3}+\dfrac{b}{2}=3a-5b-1\Leftrightarrow8a-11b=3\)

\(f\left(0\right)=2.0+m+1=m+1\)

\(\lim\limits_{x\rightarrow0^+}f\left(x\right)=\lim\limits_{x\rightarrow0^+}\dfrac{\sqrt[3]{x+1}-1}{x}=\lim\limits_{x\rightarrow0^+}\dfrac{x+1-1}{x(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1)}=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)\(f\left(0\right)=\lim\limits_{x\rightarrow0^+}f\left(x\right)\Leftrightarrow m+1=\dfrac{1}{3}\Rightarrow m=-\dfrac{2}{3}\)

Hàm liên tục với mọi \(x\ne1\)

Xét tại \(x=1\) ta có:

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(2x^2+3x\right)=2.1^2+3.1=5\)

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\left(ax+2\right)=a+2\)

\(f\left(1\right)=a+2\)

Hàm liên tục trên toàn R khi hàm liên tục tại \(x=1\)

\(\Leftrightarrow\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^+}f\left(x\right)=f\left(1\right)\)

\(\Leftrightarrow a+2=5\Rightarrow a=3\)