How can I solve limit with absolute value $\mathop {\lim }\limits_{x \to -\infty } {{x} \over {9}}|{\sin}{{6} \over {x}}|$

Question

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?

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

Answer 1

Hint:

$$\sin(f(x)) \sim f(x) \text{ as } f(x) \rightarrow 0$$

This follows directly from the Taylor expansion of $\sin(x)$. Can you take it from here?

Answer 2

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 $$ \left|\sin\frac6x\right|=-\sin\frac6x $$

Answer 3

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