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?
$$\mathop {\lim }\limits_{x \to -\infty } {{x} \over {9}}|{\sin}{{6} \over {x}}|$$

$$\lim_{x\rightarrow \infty }\frac{x}{9}\left | \sin \frac{6}{x} \right |=\frac{1}{9}\lim_{x\rightarrow \infty }x\left | \sin \frac{6}{x} \right |=\frac{1}{9}\lim_{x\rightarrow \infty }6\cos \frac{6}{x}\left | \sin \frac{6}{x} \right |'=\frac{6}{9}\lim_{x\rightarrow \infty }\cos \frac{6}{x}=\frac{2}{3}$$
