Question

Test the continuity of $f\left(x\right)$ where $f\left(x\right)=\{\begin{array}{c}{x}^2+x+1,0\le x\le 1\\ x^2+2,1<x\le 2\end{array}$

Answer 1

$\begin{array}{cc}\Rightarrow f\left(0\right)=1,f\left(1\right)=3,f\left(2\right)=6& \\ \Rightarrow {lim}_{x\to 0^{+}}=f\left(0^{+}\right)={lim}_{h\to 0}=f\left(0+h\right)={lim}_{h\to 0}{\left(0+h\right)}^2+\left(0+h\right)+1={lim}_{h\to 0}{h}^2+h=1& \\ \therefore f\left(0^{+}\right)=f\left(0\right)=1,f\left(1\right)=f\left(1^{+}\right)=3& \\ \Rightarrow {lim}_{x\to 1^{-}}f\left(1^{-}\right)={lim}_{h\to 0}f\left(1-h\right)={lim}_{h\to 0}{\left(1-h\right)}^2+\left(1-h\right)+1={lim}_{h\to 0}1-2h+h^2+1-h+1=3& \\ \Rightarrow {lim}_{x\to 1^{+}}f\left(1^{+}\right)={lim}_{h\to 0}\left(1+h\right)={lim}_{h\to 0}{\left(1+h\right)}^2+2={lim}_{h\to 0}1+2h+h^2+2=3& \left[\because f\left(2\right)=f\left(2^{-}\right)=6\right]\\ {lim}_{x\to 2^{-}}f\left(2^{-}\right)={lim}_{h\to 0}f\left(2-h\right)={lim}_{h\to 0}{\left(2-h\right)}^2+2{lim}_{h\to 0}4-2h+h^2+2=6.& \end{array}$