a) \(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{e^{3x-6}-1}{x-2}=\lim\limits_{x\rightarrow2}3\left[\dfrac{e^{3\left(x-2\right)}-1}{3\left(x-2\right)}\right]=3.1=3\)
Để hàm số \(f\left(x\right)\) liên tục tại \(x=2\) khi và chỉ khi
\(f\left(2\right)=\lim\limits_{x\rightarrow2}f\left(x\right)\)
\(\Leftrightarrow m=3\)
b) Ta thấy :
\(f'_+\left(2\right)=f'_-\left(2\right)=\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)-f\left(2\right)}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{e^{3x-6}}{x-2}-1-3}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{e^{3x-6}-1-3\left(x-2\right)}{\left(x-2\right)^2}=\lim\limits_{x\rightarrow2}\dfrac{3e^{3x-6}-3}{2\left(x-2\right)}\) (Dạng \(\dfrac{0}{0}\RightarrowĐL.L’Hospital\))
\(=\lim\limits_{x\rightarrow2}\dfrac{9e^{3x-6}}{2}=\dfrac{9}{2}\)
\(\Rightarrow\) Hàm số khả vi tại \(x=2\)