points) Evaluate $\lim _{x \rightarrow 0} \frac{2 x+\cos x-e^{2 x}}{x^{3}-3 x^{2}}$

Step 1 ：The problem is asking for the limit of a function as x approaches 0. This is a calculus problem. The function is a fraction, so we need to be careful about the denominator becoming zero. We can use L'Hopital's rule to solve this problem. L'Hopital's rule states that the limit of a ratio of two functions as x approaches a certain value is equal to the limit of the ratios of their derivatives. We can apply this rule multiple times if necessary.

Step 2 ：Let's denote the function as \(f = \frac{2x - e^{2x} + \cos(x)}{x^{3} - 3x^{2}}\)

Step 3 ：We calculate the derivative of the function, \(f' = \frac{(-3x^{2} + 6x)(2x - e^{2x} + \cos(x))}{(x^{3} - 3x^{2})^{2}} + \frac{-2e^{2x} - \sin(x) + 2}{x^{3} - 3x^{2}}\)

Step 4 ：By applying L'Hopital's rule, we find that the limit of the function as x approaches 0 is \(\frac{13}{18}\)

Step 5 ：Final Answer: The limit of the function as x approaches 0 is \(\boxed{\frac{13}{18}}\)