if (3x - 6leq f(x)leq x^{2}-3x + 3) for (xgeq0), find (lim_{x\rightarrow3}f(x)).

Explanation:

Step1: Find left - hand limit

Find $\lim_{x\rightarrow3}(3x - 6)$. Using the limit rules for linear functions, $\lim_{x\rightarrow a}(mx + b)=ma + b$. Here $m = 3$ and $b=-6$, so $\lim_{x\rightarrow3}(3x - 6)=3\times3-6=9 - 6=3$.

Step2: Find right - hand limit

Find $\lim_{x\rightarrow3}(x^{2}-3x + 3)$. Using the limit rules for polynomial functions, $\lim_{x\rightarrow a}(a_nx^n+\cdots+a_1x + a_0)=a_na^n+\cdots+a_1a + a_0$. Here $n = 2$, $a_2 = 1$, $a_1=-3$, $a_0 = 3$, so $\lim_{x\rightarrow3}(x^{2}-3x + 3)=3^{2}-3\times3 + 3=9-9 + 3=3$.

Step3: Apply Squeeze Theorem

Since $3x - 6\leq f(x)\leq x^{2}-3x + 3$ for $x\geq0$ and $\lim_{x\rightarrow3}(3x - 6)=\lim_{x\rightarrow3}(x^{2}-3x + 3)=3$, by the Squeeze Theorem, $\lim_{x\rightarrow3}f(x)=3$.

Answer:

$3$