8. $[-11$ Points] DETAILS SCALC9 1.6.021.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) \[ \lim _{h \rightarrow 0} \frac{(h-4)^{2}-16}{h} \] Need Help? Read It Watch It Master It Submit Answer
Solution
Step 1 ：The given limit is of the form \(\frac{0}{0}\) as h approaches 0. This is an indeterminate form, so we can use L'Hopital's rule to evaluate the 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. So, we need to find the derivative of the numerator and the denominator and then evaluate the limit.
Step 2 ：Let's find the derivative of the numerator and the denominator. The numerator is \((h - 4)^{2} - 16\) and the denominator is h. The derivative of the numerator is \(2h - 8\) and the derivative of the denominator is 1.
Step 3 ：Using L'Hopital's rule, the limit of the given function as h approaches 0 is equal to the limit of the derivative of the numerator divided by the derivative of the denominator. Therefore, the limit is \(-8\).
Step 4 ：Final Answer: The limit as h approaches 0 of the given function is \(\boxed{-8}\).
