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 $\mathop {\lim }\limits_{{\text{x}} \to {\text{1 }}} f(x) = 6$.