Question

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

Answer 1

Step $1:$ Finding limit at $x=0$
Given $f\left(x\right)=\{\begin{array}{cc}2x+3,& x\le 0\\ 3(x+1),& x>0\end{array}$
$L.H.L.=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(2x+3)=2\left(0\right)+3=3$
$R.H.L.=\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0}{lim}3(x+1)=3(0+1)=3$
$L.H.L.=R.H.L.=3$
$\therefore \underset{x\to 0}{lim}f\left(x\right)=3$

Step $2:$ Finding limit at $x=1$
$f\left(x\right)=\{\begin{array}{cc}2x+3,& x\le 0\\ 3(x+1),& x>0\end{array}$
For $x>0$, the function is $3(x+1)$
Therefore,
$\underset{x\to 1}{lim}f\left(x\right)=\underset{x\to 1}{lim}3(x+1)$
$=3(1+1)$
$=3\left(2\right)$
$=6$
$\therefore \underset{x\to 1}{lim}f\left(x\right)=6$

Therefore, $\underset{x\to 0}{lim}f\left(x\right)=3$ and $\underset{x\to 1}{lim}f\left(x\right)=6$.