10. [-/1 Points] DETAILS SCALC9 1.6.039.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER If $3 x-2 \leq f(x) \leq x^{2}-3 x+7$ for $x \geq 0$, find $\lim _{x \rightarrow 3} f(x)$. Need Help? Read It Watch it Master It Submit Answer

Step 1 ：The problem is asking for the limit of the function \(f(x)\) as \(x\) approaches 3. However, we don't have an explicit formula for \(f(x)\). What we do have are two inequalities that bound \(f(x)\) from above and below.

Step 2 ：The lower bound is \(3x - 2\) and the upper bound is \(x^2 - 3x + 7\).

Step 3 ：Since \(f(x)\) is squeezed between these two functions, we can use the Squeeze Theorem to find the limit. The Squeeze Theorem states that if a function \(f(x)\) is bounded by two other functions \(g(x)\) and \(h(x)\) such that \(g(x) \leq f(x) \leq h(x)\), and if the limit of \(g(x)\) and \(h(x)\) as \(x\) approaches a certain value is the same, then the limit of \(f(x)\) as \(x\) approaches that value is also the same.

Step 4 ：So, we need to find the limit of \(3x - 2\) and \(x^2 - 3x + 7\) as \(x\) approaches 3.

Step 5 ：The limit of the lower bound and the upper bound as \(x\) approaches 3 are both 7. According to the Squeeze Theorem, this means that the limit of \(f(x)\) as \(x\) approaches 3 is also 7.

Step 6 ：Final Answer: \(\boxed{7}\)