Question:
Find $\lim _{x \rightarrow 0} f(x)$ and $\lim _{x \rightarrow 1} f(x)$, where $f(x)= \begin{cases}2 x+3, & x \leq 0 \\ 3(x+1), & x>0\end{cases}$
Solution:
The given function is
$f(x)= \begin{cases}2 x+3, & x \leq 0 \\ 3(x+1), & x>0\end{cases}$
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}[2 x+3]=2(0)+3=3$
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} 3(x+1)=3(0+1)=3$
$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} f(x)=3$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 3(x+1)=3(1+1)=6$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 3(x+1)=3(1+1)=6$
$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x)=6$