(2) $\lim _{x_{-} \rightarrow 3} \frac{9-x^{2}}{3-x}$

Step 1 ：We are given the limit \(\lim _{x_{-} \rightarrow 3} \frac{9-x^{2}}{3-x}\).

Step 2 ：As x approaches 3, the given limit is of the form \(\frac{0}{0}\). This is an indeterminate form.

Step 3 ：We can use L'Hopital's rule to solve this limit. L'Hopital's rule states that if the limit of a function is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then the limit of that function is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

Step 4 ：Let's find the derivatives of the numerator and the denominator. The derivative of the numerator \(9 - x^{2}\) is \(-2x\) and the derivative of the denominator \(3 - x\) is \(-1\).

Step 5 ：Applying L'Hopital's rule, we find that the limit is equal to the limit of the ratio of the derivatives, which is \(\frac{-2x}{-1} = 2x\).

Step 6 ：Substituting x = 3, we find that the limit is 6.

Step 7 ：Final Answer: The limit of the function as x approaches 3 from the left is \(\boxed{6}\).