Chắc đề bài là xét tính liên tục của hàm số?
e.
\(\lim\limits_{x\rightarrow5}f\left(x\right)=\lim\limits_{x\rightarrow5}\dfrac{\sqrt[]{4x+16}+4-2x}{x^2-5x}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\left(\sqrt[]{4x+16}-6\right)+10-2x}{x^2-5x}=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{\left(\sqrt[]{4x+16}-6\right)\left(\sqrt[]{4x+16}+6\right)}{\sqrt[]{4x+16}+6}-2\left(x-5\right)}{x\left(x-5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{4\left(x-5\right)}{\sqrt[]{4x+16}+6}-2\left(x-5\right)}{x\left(x-5\right)}=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{4}{\sqrt[]{4x+16}+6}-2}{x}\)
\(=\dfrac{\dfrac{4}{\sqrt[]{4.5+16}+6}-2}{5}=-\dfrac{1}{3}\)
\(f\left(5\right)=-\dfrac{1}{3}\)
\(\Rightarrow\lim\limits_{x\rightarrow5}f\left(x\right)=f\left(5\right)\)
Hàm liên tục tại \(x_0=5\)
f.
\(\lim\limits_{x\rightarrow-1}f\left(x\right)=\lim\limits_{x\rightarrow-1}\dfrac{x^3+x+2}{x^3+1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^2-x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{x^2-x+2}{x^2-x+1}\)
\(=\dfrac{1+1+2}{1+1+1}=\dfrac{4}{3}\)
Và \(f\left(-1\right)=\dfrac{4}{3}\)
\(\Rightarrow\lim\limits_{x\rightarrow-1}f\left(x\right)=f\left(-1\right)\)
Hàm liên tục tại \(x_0=-1\)
h.
\(\lim\limits_{x\rightarrow-3}f\left(x\right)=\lim\limits_{x\rightarrow-3}\dfrac{-2x^2-3x+9}{\sqrt{6-x}-3}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{\left(3-2x\right)\left(x+3\right)\left(\sqrt{6-x}+3\right)}{\left(\sqrt{6-x}-3\right)\left(\sqrt{6-x}+3\right)}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{\left(3-2x\right)\left(x+3\right)\left(\sqrt{6-x}+3\right)}{-\left(x+3\right)}\)
\(=\lim\limits_{x\rightarrow-3}\dfrac{\left(3-2x\right)\left(\sqrt{6-x}+3\right)}{-1}=-54\)
Và \(f\left(-3\right)=-54\)
\(\Rightarrow\lim\limits_{x\rightarrow-3}f\left(x\right)=f\left(-3\right)\)
Hàm liên tục tại \(x_0=-3\)
