How can I solve limit with absolute value $\mathop {\lim }\limits_{x \to -\infty } {{x} \over {9}}|{\sin}{{6} \over {x}}|$ I try to evaluate this limit by L'Hopitals rule, but I don't know how affect limit absolute value. Can someone give me some advice please?
$$\mathop {\lim }\limits_{x \to -\infty } {{x} \over {9}}|{\sin}{{6} \over {x}}|$$

L'Hôpital is not a magical thing to be used all the time. Thinking is advantageous. 
If you want to consider $x\to-\infty$, you will only be dealing with $x<0$. So the numbers $6/x$ are going to be negative and close to zero. In that zone, the sine is negative, so 
$$
\left|\sin\frac6x\right|=-\sin\frac6x
$$
