u can apply lhopital till your ink runs dry so long as the conditions of indeterminacy are true. in your case, one more time will do it. (YOU SHOULD KNOW THAT!)
This may or not employ L'Hospital's, but here it is:
Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...
which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.