Find the derivative: lim x-->0 (sin3x)/(5x^3-4x)

Question

I'm pretty lost on how to begin this problem. 

Answer 1

This problem is confusing.  I'm guessing the "Find the derivative" part is just a hint for you to use L'hopital's rule to evaluate the limit (since it tends to 0/0) by taking the derivative of the top and bottom of the quotient to get 

 

lim(x-->0) (3cos(3x)) / ( 15x^2 - 4)  

 

which evaluates to 3 / (-4) = -3/4