Question

\(f\left(x\right)=\left\{{}\begin{matrix}\dfrac{x^2-3x+2}{x^3+8}\left(x\ne-2\right)\\mx+1\left(x=-2\right)\end{matrix}\right.\)

tìm m để hàm số gián đoạn tại \(x=-2\)

Answer 1

\(f\left(-2\right)=-2m+1\)

\(\lim\limits_{x\rightarrow-2^+}f\left(x\right)=\lim\limits_{x\rightarrow-2^+}\dfrac{x^2-3x+2}{x^3+8}=\lim\limits_{x\rightarrow-2^+}\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\lim\limits_{x\rightarrow-2^+}\dfrac{x-1}{x^2-2x+4}=\dfrac{-2-1}{4-2.\left(-2\right)+4}=-\dfrac{1}{4}\)

\(f\left(-2\right)\ne\lim\limits_{x\rightarrow-2^-}f\left(x\right)\Leftrightarrow-2m+1\ne-\dfrac{1}{4}\Leftrightarrow m\ne\dfrac{5}{8}\)