# trigonometry limit finding using squeeze thereom didn't work out

I have the function $$f(x)=\frac{1}{2^x}-3x+2, \quad x\in\mathbb{R}$$ and I want to find the $$\lim_{x\to -\infty}{\frac{\sin(f(x))}{x}}.$$ I have that $$\lim_{x\to -\infty}{f(x)}=+\infty,$$ so I am thinking of using the squeeze theorem to find $$\lim_{x\to -\infty}{\frac{\sin(f(x))}{x}},$$ but this wasn't work for me. Any ideas? (I don't want to use de L'Hospital Rule)

• Hint: $\sin$ is bounded. – Michael Burr Feb 6 '18 at 14:10
• Start with $-1 \leq \sin(f(x)) \leq 1$. – Jonathan Feb 6 '18 at 14:11

$$\left|\frac{\sin(f(x))}{x}\right| \le \left|\frac{1}{x}\right|$$
and this directly leads to $0$.