1.
\(\lim\limits_{x\rightarrow0}\dfrac{e^x-1}{x}=1=f\left(0\right)\) nên hàm liên tục tại \(x=0\)
Khi đó:
\(f'\left(0\right)=\lim\limits_{x\rightarrow0}\dfrac{f\left(x\right)-f\left(0\right)}{x-0}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{e^x-1}{x}-1}{x}=\lim\limits_{x\rightarrow0}\dfrac{e^x-1-x}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{e^x-1}{2x}=\dfrac{1}{2}\)
2.
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\left(x^3-3x\right)=1^3-3.1=-2\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(3-x\right)=3-1=2\)
\(\Rightarrow\lim\limits_{x\rightarrow1^+}f\left(x\right)\ne\lim\limits_{x\rightarrow1^-}f\left(x\right)\)
\(\Rightarrow\) Hàm gián đoạn tại \(x=1\) nên ko tồn tại \(f'\left(1\right)\)
