$\underset{x→0}{\lim}\frac{\sin(6x^2)}{ln\,\cos(2x^2-x)}$ is equal to:

-12

$\underset{x→0}{\lim}\frac{\sin(6x^2)}{ln(\cos(2x^2-x))}=\underset{x→0}{\lim}\frac{\cos(6x^2)×12x}{\frac{-\sin(2x^2-x)}{\cos(2x^2-x)}.(4x-1)}$ [Applying L Hospitals Rule]

$=\underset{x→0}{\lim}\frac{1×12×x}{\frac{+\sin(2x^2-x)}{1}×(1)}=12\underset{x→0}{\lim}\frac{(2x^2-x)}{\sin(2x^2-x)×\frac{x}{(2x^2-x)}}=12\underset{x→0}{\lim}\frac{1}{2x-1}=-12$