Find the following limit.
\[
\lim _{x \rightarrow 3}\left((3-x)^{2} \cos \left(\frac{1}{3-x}\right)+3\right)
\]
If the limit does not exist, click on "Does Not Exist."

Step 1 ：We are given the function \(f = (3 - x)^2\cos\left(\frac{1}{3 - x}\right) + 3\) and we are asked to find the limit as \(x\) approaches 3.

Step 2 ：The function is continuous at all points except \(x=3\). At \(x=3\), the function is undefined due to division by zero in the cosine function.

Step 3 ：However, we can still find the limit as \(x\) approaches 3 by using L'Hopital's rule, which states that if the limit of a function as \(x\) approaches a certain value results in an indeterminate form (0/0 or ∞/∞), 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 ：In this case, we can rewrite the function as a fraction with the numerator as \((3-x)^2\cos\left(\frac{1}{3-x}\right)\) and the denominator as 1, and then apply L'Hopital's rule.

Step 5 ：The derivative of the function is \((2x - 6)\cos\left(\frac{1}{3 - x}\right) - \sin\left(\frac{1}{3 - x}\right)\).

Step 6 ：The limit of the derivative of the function as \(x\) approaches 3 is a range of possible values. This means that the limit of the function as \(x\) approaches 3 does not exist in a traditional sense, because it does not approach a specific value. Instead, it oscillates between -1 and 1.

Step 7 ：\(\boxed{\text{The limit does not exist.}}\)