$\lim _{\lim \rightarrow 4} \frac{x^{2}-16}{x-4}$

Step 1 ：We are given the limit \(\lim _{x \rightarrow 4} \frac{x^{2}-16}{x-4}\).

Step 2 ：This limit is of the form \(\frac{0}{0}\) when x approaches 4, which 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 ：We find the derivative of the numerator, which is \(2x\), and the derivative of the denominator, which is \(1\).

Step 5 ：We then find the limit of the ratio of these two derivatives as x approaches 4, which is \(8\).

Step 6 ：Final Answer: The limit of the given function as x approaches 4 is \(\boxed{8}\).