We have the function,
$f(x) = \left\{ \begin{gathered}
2x + 3x \leqslant 0 \hfill \\
3\left( {x + 1} \right)x > 0 \hfill \\
\end{gathered} \right.$
Evaluating $\mathop {\lim }\limits_{x \to 0} f(x)$,
When, $\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to 0} (2x + 3)$
$ = 2(0) + 3$
$ = 0 + 3$
$ = 3$
And when, $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to 0} 3(x + 1)$
$ = 3(0 + 1)$
$ = 3(1)$
$ = 3$
Thus, $\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to 0} f(x) = 3$.
Now, evaluating $\mathop {\lim }\limits_{x \to 1} f(x)$,
When, $\mathop {\lim }\limits_{x \to {1^ – }} f(x) = \mathop {\lim }\limits_{x \to 1} 3(x + 1)$
$ = 3(1 + 1)$
$ = 3(2)$
$ = 6$
And when, $\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to 1} 3(x + 1)$
$ = 3(1 + 1)$
$ = 3(2)$
$ = 6$
Thus, $\mathop {\lim }\limits_{x \to {1^ – }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{x \to 1} f(x) = 6$
Therefore, $\mathop {\lim }\limits_{x \to 0} f(x) = 3$ and $\mathop {\lim }\limits_{{\text{x}} \to {\text{1 }}} f(x) = 6$.