Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x),$ where $f(x) = \begin{cases} 2x+3, & x \leq 0 \\ 3(x+1), & x > 0 \end{cases}$

(N/A) The given function is $f(x) = \begin{cases} 2x+3, & x \leq 0 \\ 3(x+1), & x > 0 \end{cases}$
For $\mathop {\lim }\limits_{x \to 0} f(x)$:
Left-hand limit: $\mathop {\lim }\limits_{x \to 0^-} f(x) = \mathop {\lim }\limits_{x \to 0} (2x+3) = 2(0)+3 = 3$
Right-hand limit: $\mathop {\lim }\limits_{x \to 0^+} f(x) = \mathop {\lim }\limits_{x \to 0} 3(x+1) = 3(0+1) = 3$
Since the left-hand limit equals the right-hand limit,$\mathop {\lim }\limits_{x \to 0} f(x) = 3$.
For $\mathop {\lim }\limits_{x \to 1} f(x)$:
Since $x=1$ is in the domain $x > 0$,we use the function $f(x) = 3(x+1)$.
$\mathop {\lim }\limits_{x \to 1} f(x) = \mathop {\lim }\limits_{x \to 1} 3(x+1) = 3(1+1) = 6$.