Question

Find $\underset{x\to 0}{lim}f\left(x\right)$ and $\underset{x\to 1}{lim}f\left(x\right)$ where $f\left(x\right)=$ $\{\begin{array}{c}2x+3,x\le 0\\ 3(x+1),x>0\end{array}$

Answer 1

$f\left(x\right)=\{\begin{array}{cc}2x+3,& x\le 0\\ 3(x+1),& x>0\end{array}$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}[2x+3]=2\left(0\right)+3=3$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}3(x+1)=3(0+1)=3$

$\therefore \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}f(x=3)$

$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}3(x+3)=3(1+1)=6$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1}{lim}3(x+1)=3(1+1)=6$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}3(x+1)=3(1+1)=6$

$\therefore \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)=6$